General Chemistry 402101-4.

General Chemistry

الاختبارات و المواعيد المهمة
العملي 30 درجة النظري 70 درجة 20 درجة نصفي ألاسبوع ال10 40 درجة اختبار نهائي الأسبوع 18 10 درجات كويز الأسبوع 13 أو 14 10 درجات (اختبار) الاسبوع 16 أو 17 20 درجة (اجراء التجارب) 2

Chapter 1 Matter

The study of matter and the changes it undergoes
First lecture What is Chemistry? The study of matter and the changes it undergoes

Matter mixture pure substance compound element
First lecture Matter mixture pure substance compound element Separation by physical methods Separation by chemical methods homogeneous heterogeneous Any thing that occupies space and has space A combination of two or more substances in which the substances retain their distinct identities Has a definite composition and distinct properties composed of two different elements or more chemically united in fixed proportions. The composition is not uniform The composition is the same throughout cannot be separated into simpler substances by chemical means

NaCl Salt water Iron sugar air helium water salad
First lecture NaCl Salt water Iron sugar air helium water salad compound homogeneous mixture element compound homogeneous mixture element compound heterogeneous mixture compound element homogeneous mixture heterogeneous

Molecules containing different two atoms or more
First lecture Compound Molecules containing different two atoms or more Element Atoms Molecules C, Al, Ti H2O, H2SO3 Polyatomic Diatomic S8, O3 O2 , H2

First lecture Matter States The difference between the states is the distance between the molecules.

Physical Chemical Matter properties color, mass, size reactivity,
First lecture Matter properties Chemical Physical color, mass, size reactivity, flammability is a property when the matter undergoes a chemical change or reaction Can be measured and observed without changing the composition or identity of a substances

Measurable properties of matter
First lecture Measurable properties of matter extensive intensive depends on how much matter is being considered Does not depend on how much matter is being considered Mass volume Density temperature How can these properties be measured ?

International system of units
First lecture Measurement SI Units International system of units Symbol Name of unit Base Quantity m meter Length Kg Kilogram Mass s Second Time A Ampere Electrical current K Kelvin Temperature mol Mole Amount of substance cd candela Luminous intensity

Prefixes Used with SI Units
First lecture Prefix Symbol Multiple of Base Unit Giga G 1,000,000,000 or 109 Mega M 1,000,000 or 106 kilo k 1,000 or 103 deci d 0.1 or 10-1 centi c 0.01 or 10-2 milli m 0.001 or 10-3 micro or 10-6 nano n 10-9 pico p 10-12 Femto f 10-15

Mass and weight Newton (N)
First lecture Mass and weight What is the difference between mass and weight? Mass: is a measure of amount of matter in an object Weight: is the force that gravity exerts on an object Newton (N)

Volume Volume – SI derived unit for volume is cubic meter (m3)
First lecture Volume Volume – SI derived unit for volume is cubic meter (m3) 1 cm3 = (1 x 10-2 m)3 = 1 x 10-6 m3 1 cm3 = 1 mL 1 L = 1000 mL = 1000 cm3 = 1 dm3

Dimensional Analysis Method of Solving Problems
First lecture Dimensional Analysis Method of Solving Problems How many mL are in 1.63 L? Conversion Unit 1 L = 1000 mL 1L 1000 mL 1.63 L x = 1630 mL 1L 1000 mL 1.63 L x = L2 mL

Density g/cm3 for solids g/ml for liquids g/L for gases
First lecture Density Density is defined as the mass per unit volume. density = mass/volume d = m V S.I. units for density = kg/m3 g/cm3 for solids g/ml for liquids g/L for gases

A piece of platinum metal with a density of 21.5 g/cm3 has a volume of
First lecture A piece of platinum metal with a density of 21.5 g/cm3 has a volume of 4.49 cm3. What is its mass? d = m V m = d x V m = 21.5 g/cm3 x 4.49 cm3 = 96.5 g

Temperature T(in Kelvin) = T(in Celsius) + 273.15 Temperature scales
First lecture Temperature scales Fahrenheit oF Celsius oC Kelvin K 0F = ( x 0C )+ 32 9 5 32 0F = 0 0C 212 0F = 100 0C 273 K = 0 0C 373 K = 100 0C T(in Kelvin) = T(in Celsius)

Precision and Accuracy
First lecture Precision and Accuracy Student A Student B Student C 1.964 g g g 1.978 g g g Average g g g The true mass of object= g Precision: how close a set of measurements are to each other (reproducibility). Accuracy: how close your measurements are to the true value.

Significant Figures Any digit that is not zero is significant
First lecture Any digit that is not zero is significant 1.234 kg 4 significant figures Zeros between nonzero digits are significant 606 m 3 significant figures Zeros to the left of the first nonzero digit are not significant 0.08 L 1 significant figure If a number is greater than 1, then all zeros to the right of the decimal point are significant 2.0 mg 2 significant figures If a number is less than 1, then only the zeros that are at the end and in the middle of the number are significant g significant figures

First lecture Exact Numbers Numbers from definitions or numbers of objects are considered to have an infinite number of significant figures The average of three measured lengths; 6.64, 6.68 and 6.70? 3 = = 6.67 = 7 Because 3 is an exact number

Scientific Notation N x 10n The number of atoms in 12 g of carbon:
First lecture N x 10n The number of atoms in 12 g of carbon: 602,200,000,000,000,000,000,000 n is a positive or negative integer 6.022 x 1023 N is a number between 1 and 10 The mass of a single carbon atom in grams: 1.99 x 10-23

First lecture How many significant figures are in each of the following measurements? Tip: start to count the sig. fig. from the left when you see a non zero number until the end of the number.

Significant Figures: Addition & Subtraction
First lecture Significant Figures: Addition & Subtraction If addition or subtraction: 1- must have same power before addition or subtraction 2- sig. fig. in the answer is as the smaller digits after decimal point

Significant Figures: Multiplication & Division
First lecture Significant Figures: Multiplication & Division If multiplication or division: 1- add exponent for multiplication or subtract exponent for division write the answer with the smaller sig. fig.

Questions Question 1 Question 3 Question 4 Question 2 First lecture
Which of the following is an example of a physical property? A) combustibility B) corrosiveness C) explosiveness D) density E) A and D Question 2 Which of the following represents the greatest mass? A) 2.0 x 103 mg B) 10.0 dg C) kg D) 1.0 x 106 μg E) 3.0 x 1012 pg Question 3 Convert 240 K and 468 K to the Celsius scale. A) 513oC and 741oC B) -59oC and 351oC C) -18.3oC and 108oC D) -33oC and 195oC Question 4 Calculate the volume occupied by 4.50 X 102 g of gold (density = 19.3 g/cm3). A) 23.3 cm3 B) 8.69 x 103 cm C) 19.3 cm3 D) 450 cm3

Questions Question 5 Question 6 Question 7 Question 8 First lecture
The melting point of bromine is -7oC. What is this melting point expressed in oF? A) 45oF B) -28oF C) -13oF D) 19oF E) None of these is within 3oF of the correct answer. Question 6 How many significant figures are there in the measurement g? A) 6 B) 5 C) 4 D) 3 E) 2 Question 7 How many significant figures should you report as the sum of ? A) 3 B) 5 C) 7 D) 6 E) 4 Question 8 How many significant figures are there in the number g? A) 8 B) 7 C) 6 D) 5

Questions Question 11 Question 9 Question 10 First lecture
The value of 345 mm is a measure of A) temperature B) density C) volume D) distance E) Mass Question 10 The measurement m, expressed correctly using scientific notation, is 0.43 x m 4.3 x 10-6 4.3 x 10-7 4.3 x 10-5 Question 11 A laboratory technician analyzed a sample three times for percent iron and got the following results: 22.43% Fe, 24.98% Fe, and % Fe. The actual percent iron in the sample was 22.81%. The analyst's A) precision was poor but the average result was accurate. B) accuracy was poor but the precision was good. C) work was only qualitative. D) work was precise. E) C and D.

2 Atomic, Molecular Weights & Moles Calculations in Chemical Equations
Second lecture Atomic, Molecular Weights & Moles Calculations in Chemical Equations Chapter 2

Molar Mass (Atomic weight Aw)
Second lecture Atomic Mass The mass of an atom in atomic mass units (amu) 6 Atomic number C 12.01 Atomic mass The atomic mass of elements is relative to a standard atom 12 C (6 protons, 6 neutrons) Molar Mass (Atomic weight Aw) The mass of an element atoms per one mole (g/mol) = Atomic Mass numerically

Second lecture Mole (mol) The amount of a substance that contains as many elementary particles (atoms, molecules or ions), where each mole has number of × particles. 1 mole= × particles = Avogadro’s number Na 1 mol Al = 6.02 × atoms 1 mol CO2 = 6.02× 1023 molecules 1 mol NaCl = 6.02× Na+ ions = 6.02× Cl- ions The number of atoms in exactly 12 g of 12C is one mole

Molar Mass ( Atomic weight Aw):
Second lecture Molar Mass ( Atomic weight Aw): mass (weight) of 1 mole of atoms in grams 1 mol C atoms = g Aw of C = * g/mol 1 mol Cl atoms = g Aw of Cl = * g/mol 1 mol Fe atoms = g Aw of Fe = * g/mol *( get from periodic table) Think: What is the difference between the mass and weight?

Molar Mass ( Molecular weight Mw):
Second lecture Molar Mass ( Molecular weight Mw): The sum of atomic weights of 1 mol of the molecule Mw of 1 mol of H2O = 2 (Aw of H) + Aw of O = (2× 1.008) + 16 = g/mol

Learning check C2H6 What are the molecular weights of the following:
Second lecture Learning check What are the molecular weights of the following: C2H6 N2O4 C8H18O4N2S Al2(CO3)3 MgSO4.7H2O

Second lecture Number of moles (n) Remember: No. of particles = No. of moles × Avogadro's number

Example n = wt/Aw = 6.46/4.003 = 1.61 mol He
Second lecture Example Helium (He) is a valuable gas used in industry, low temperature research, deep-sea diving tanks, and balloons. How many moles of He atoms are in 6.46 g of He? 1 mol of He = g i.e. Aw = g/mol n = wt/Aw = 6.46/4.003 = 1.61 mol He

Second lecture Example

Learning check What is the number of moles in 21.5 g CaCO3?
Second lecture Learning check What is the number of moles in 21.5 g CaCO3? What is the mass in grams of 0.6 mol C4H10? How many atoms of Cu are present in 35.4 g of Cu? 38

Percent Composition of Compounds
Second lecture Percent Composition of Compounds Mass percent (weight percent) of each element in a compound.

Calculate the mass percent of each element in ethanol (C2H5OH) ?
المحاضرة الأولى Example Calculate the mass percent of each element in ethanol (C2H5OH) ? Mass of 1 mol (molar mass) of C2H5OH = = g/mol Mass percent of C = 2 x g/mol g/mol x 100 = % (4 sf) Mass percent of H = 6 x g/mol g/mol x 100 = % (4 sf) Mass percent of O = 1 x g/mol g/mol x 100 = % (4 sf) Total mass = =100%

CH2O C6H12O6 Percent composition Determining the Formula of a Compound
المحاضرة الأولى Percent composition Determining the Formula of a Compound empirical formula molecular formula CH2O C6H12O6 , 𝑥 = 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑀𝑓 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝐸𝑓 Molecular formula = (Empirical formula)𝒙

Firstly: Determination of the empirical formula:
المحاضرة الأولى Determine the empirical and molecular formulas for a compound that gives the following mass: % N % O Molar mass = g/mol Firstly: Determination of the empirical formula: Assume we have a g sample WN = g, WO = g 1- Calculate number of moles of each element in 100 g of a compound nN = (30.46 g)/(14.01 g mol-1) = mol N round to 2sf nO = (69.54 g)/(16.00 g mol-1) = mol O round to 2sf

Secondly: Determination of molecular formula
المحاضرة الأولى 2 -Determine the molar ratio by dividing by the smallest subscript mole nN : nO : : The empirical formula NO2 Secondly: Determination of molecular formula 1- Calculate the empirical formula mass Empirical molar mass = (1 x 14.01) + (2 x 16.00) = g 2- Calculate the ratio between the molar mass and the empirical molar mass: 𝑥= 𝑀𝑓 𝐸𝑓 = (92.02 g)/(46.01 g) =2 Molecular formula = (NO2)𝒙 = (NO2)2 = N2O4

Second lecture Learning check 44

Second lecture Chemical Reactions Reactants Products A process in which one or more substances is changed into one or more new substances. 2H2 (g) + O2 (g) 2H2O (l) 2HgO (s) 2Hg (l) + O2 (g) 45

Reactants (starting materials) Products ( materials formed)
Second lecture Chemical Equations It is a way to represent the chemical reaction. It shows us: The chemical symbols of reactants and products The physical states of reactants and products– (s), (l), (g), (aq) Balanced equation (same number of atoms on each side) 2H2 (g) + O2 (g) 2H2O (l) Reactants (starting materials) Products ( materials formed) 46

Balancing Chemical Equations
Second lecture Balancing Chemical Equations The number of atoms of each element must be the same on both sides of the equation. C2H6 + O2 CO2 + H2O C2H6 + 7/2O2 2CO2 + 3H2O 2C2H6 + 7O2 4CO2 + 6H2O Reactants Products 2 C 1 C 6 H 2 H 2 O 3 O Reactants Products 4 C 12 H 14 O 47

Balance the following equations:
Second lecture Balance the following equations: 48

Second lecture Stoichiometry The quantitative study of reactants and products in a chemical reaction CH4 + O2  CO2 + H2O 49

The Mole Method: N2 (g) + 3 H2 (g)  2NH3 (g)
Second lecture The Mole Method: Stoichiometric coefficients in a chemical equation can be interpreted as the number of moles of each substance. N2 (g) + 3 H2 (g)  2NH3 (g) N2 1 mole 6.022 ×1023 molecules H2 3 mole 3×6.022 ×1023 molecules NH3 2 mole 2×6.022 ×1023 molecules 50

Mole Ratios N2 (g) + 3 H2 (g)  2NH3 (g)
Second lecture Mole Ratios N2 (g) + 3 H2 (g)  2NH3 (g) Recall that the coefficient on N2 is 1 but is not explicitly written in the reaction Coefficients: N2 = 1 H2 = 3 NH3 = 2 Using the coefficients we can write mole ratios Definition: mole ratio gives the relative amounts of reactants and products

Second lecture N2 (g) + 3 H2 (g)  2NH3 (g) 52

MOLE to MOLE Stoichiometry
Second lecture MOLE to MOLE Stoichiometry N2 (g) + 3 H2 (g)  2NH3 (g) 53

Example Si (s) + 2Cl2 (g) SiCl4 (l)
Second lecture Example Silicon tetrachloride (SiCl4) can be prepared by heating Si in chlorine gas: Si (s) + 2Cl2 (g) SiCl4 (l) In one reaction, mole of SiCl4 is produced. How many moles of molecular chlorine were used in the reaction? Cl2 SiCl4 2 mol 1 mol ?? mol n of Cl2 used = (0.507 molSiCl4 ×( 2 molCl2 /1 molSiCl4) = 1.01 molCl2 54

Second lecture Example If 85.0 g of CH4 is consumed by a person over a certain period, what is the mass of CO2 produced? CH4 + 2O2  CO2 + 2H2O Convert 85.0 g to moles: n (CH4) = wt/Mw = (85.0/16.04) = 5.30 mol(CH4) CH4 + 2O2  CO2 + 2H2O 1 mole of CH mole of CO2 5.30 moles ?? n (CO2) = 5.30 mol(CH4) × (1mol (CO2) /1mol (CH4) ) = 5.30 mol(CO2) wt (CO2) = n × Mw = 5.30 mol(CO2)× (44.01g (CO2) / 1mol (CO2) ) = g = 2.33 × 102 g (CO2) 55

Second lecture You can use this version of the mole map to solve stoichiometry problems. 56

Questions Question 1 Question 3 Question 4 Question 2 Second lecture
Determine the number of moles of aluminum in kg of Al. A) x 1023 mol B) x 103 mol C) mol D) mol E) x 10-3 mol Question 2 How many phosphorus atoms are there in 2.57 g of P? A) 4.79 x 1025 B) 1.55 x 1024 C) 5.00 x 1022 D) 8.30 x 10-2 E) 2.57 Question 3 One mole of H2 A) contains 6.0 x 1023 H atoms B) contains 6.0 x 1023 H2 molecules C) contains 1 g of H2 D) is equivalent to 6.02 x 1023 g of H2 E) None of the above Question 4 How many oxygen atoms are present in 5.2 g of O2? A) 5.4 x atoms B) 9.8 x 1022 atoms C) 2.0 x 1023 atoms D) 3.1 x 1024 atoms E) 6.3 x 1024 atoms 57

Questions Question 5 Question 7 Question 6 Second lecture
How many protons and neutrons are in sulfur-33? A) 2 protons, 16 neutrons B) 16 protons, 31 neutrons C) 16 protons, 17 neutrons D) 15 protons, 16 neutrons Question 6 What is the mass of 5.45 x 10-3 mol of glucose, C6 H12O6? A) g B) 982 g C) 3.31 x 104 g D) g E) None of the above. Question 7 Determine the mass percent of iron in Fe4[Fe(CN)6] 3. A) 45% Fe B) 26% Fe C) 33% Fe D) 58% Fe E) None of the above Question 8 What is the coefficient for CO2 when the following chemical equation is properly balanced using the smallest set of whole numbers? C4H10 + O2 ----> CO2 + H2O A) 1 B) 4 C) 6 D) 8 E) 12 58

Questions Question 9 Second lecture
Calculate the number of moles of H2O formed when mole of Ba(OH)2 is treated with mol of HClO3 according to the chemical reaction shown below. Ba(OH)2+2HClO3 ----> Ba(ClO3)2 + 2 H2O A) 1.00 mol B) mol C) mol D) mol E) mol

3 Chemical Reactions in Solutions: Acids & Bases
Third lecture Chemical Reactions in Solutions: Acids & Bases Concentrations & pH Calculations Chapter 3

Solutions Solution: a homogeneous mixture of two or more substances
Third lecture Solutions Solution: a homogeneous mixture of two or more substances Solute: a substance that is being dissolved (smaller amount) Solvent: a substance which dissolves a solute (larger amount) Solute particle

Concentrations v/v w/w w/v Concentrations Molar concentrations
Third lecture Concentrations The concentration of a solution is the amount of solute present in a given quantity of a solvent or solution. Concentrations Molar concentrations Percentages Mole fraction Molarity Molality Normality v/v w/w w/v

Molarity The number of moles of solute dissolved in one liter of
Third lecture Molarity The number of moles of solute dissolved solution. in one liter of What is the unit of molarity? What is the relationship between weight and molarity?

Third lecture Example A solution has a volume of 2.0 L and contains 36.0 g of glucose (C6H12O6). If the molar mass of glucose is 180 g/mol, what is the molarity of the solution? No. of mol of glucose = wt (g) / Mw (g/mol) = 36.0 g /180 g/mol = 0.2 mol M = n (mol) / V (L) = 0.2 mol /2.0 L = 0.1 mol/L

Third lecture Molality The number of moles of solute dissolved in one kilogram of solvent Molality (m) Molarity (M) m = moles of solute mass of solvent (kg) M = moles of solute liters of solution

Third lecture Example What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is g/mL? m = moles of solute mass of solvent (kg) Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol 927 g of solution (1000 mL x g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = kg m = moles of solute mass of solvent (kg) = 5.86 moles C2H5OH 0.657 kg solvent = 8.92 m 66

Third lecture Learning check What is the concentration of a solution in mol/L when 80 g of calcium carbonate, CaCO3, is dissolved in 2 L of solution? How many liters of 0.25 M NaCl solution must be measured to obtain 0.1 mol of NaCl? A student needs to prepare 250 ml of 0.1 M of Cd(NO3)2 solution. How many grams of cadmium nitrate are required?

Questions Question 3 Question 4 Third lecture Question 1
Molarity is the number of …… of solute dissolved Solution Grams Milliliter Second moles Question 2 Molality is the number of moles of ……. dissolved in 1kg solvent Solvent Solute acid Question 3 Molarity is the number of moles of solute dissolved 1 …… of the Solution Grams Liter Second moles Question 4 A solution has a volume of 2.0 L and contains 36.0 g of glucose (C6H12O6). If the molar mass of glucose is 180 g/mol, what is the molarity of the solution 1.0 1.00 0.1 0.01

Questions Question 5 Question 7 Question 6 Third lecture B) 2 C) 2.5
How many liters of 0.25 M NaCl solution must be measured to obtain 0.1 mol of NaCl A) 1 B) 2 C) 2.5 D) 3.5 Question 6 What is the concentration of a solution in mol/L when 80 g of calcium carbonate, Ca(CO3)2, is dissolved in 2 L of solution? (Molecular weight of Ca(CO3)2 = 100g/mol A) 0.4 B) 4 C) 0.004 D) 1 Question 7 A student needs to prepare 250 ml of 0.1 M of Cd(NO3)2 solution. How many grams of cadmium nitrate are required? (Molecular weight of Cd(NO3)2 = 236 g/mol A) 5.9 B) 5.1 C) 5.4 D) 5.6

Fourth lecture Chapter 4 Chemical Equilibrium

Physical equilibrium is between two states of the same substance
Fourth lecture Equilibrium Equilibrium is a state in which there are no observable changes as time goes by Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant Chemical equilibrium Physical equilibrium N2O4 (g) 2NO2 (g) H2O (l) H2O (g) Physical equilibrium is between two states of the same substance N2O4 Colorless Equilibrium state NO2 Dark-brown

N2O4 (g) 2NO2 (g) Start with NO2 Start with N2O4 Start with NO2 & N2O4
Fourth lecture N2O4 (g) NO2 (g) equilibrium equilibrium equilibrium Start with NO2 Start with N2O4 Start with NO2 & N2O4

Equilibrium Constant K
Fourth lecture Equilibrium Constant K N2O4 (g) NO2 (g) Kp = NO2 P 2 N2O4 P Kc = [NO2]2 [N2O4] K = [NO2]2 [N2O4] = 4.63 x 10-3 aA + bB cC + dD K = [C]c[D]d [A]a[B]b Law of Mass Action

Equilibrium Position K>>1 K<<1
Fourth lecture Equilibrium Position K>>1 Products are favored at equilibrium (the equilibrium lie to the right) K<<1 Reactants are favored at equilibrium (the equilibrium lie to the left)

Relation between Kc and Kp
Fourth lecture Relation between Kc and Kp N2O4 (g) NO2 (g) Kp = NO2 P 2 N2O4 P Kc = [NO2]2 [N2O4] In most cases Kc  Kp aA (g) + bB (g) cC (g) + dD (g) Kp = Kc (RT)Dn Dn = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)

Homogeneous Equilibrium
Fourth lecture Homogeneous Equilibrium Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq) [CH3COO-][H3O+] [CH3COOH][H2O] Kc = ′ [H2O] = constant [CH3COO-][H3O+] [CH3COOH] = Kc [H2O] ′ Kc = General practice not to include units for the equilibrium constant.

Fourth lecture The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = M, [Cl2] = M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. CO (g) + Cl2 (g) COCl2 (g) [COCl2] [CO][Cl2] = 0.14 0.012 x 0.054 Kc = = 220 Kp = Kc(RT)Dn Dn = 1 – 2 = -1 R = T = = 347 K Kp = 220 x ( x 347)-1 = 7.7

The equilibrium constant Kp for the reaction:
Fourth lecture The equilibrium constant Kp for the reaction: is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = atm and PNO = atm? 2NO2 (g) NO (g) + O2 (g)

Heterogeneous Equilibrium
Fourth lecture Heterogeneous Equilibrium Heterogenous equilibrium applies to reactions in which reactants and products are in different phases CaCO3 (s) CaO (s) CO2 (g) [CaO][CO2] [CaCO3] Kc = ′ [CaCO3] = constant [CaO] = constant [CaCO3] [CaO] Kc x ′ Kc = [CO2] = The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

Consider the following equilibrium at 295 K:
Fourth lecture Consider the following equilibrium at 295 K: The partial pressure of each gas is atm. Calculate Kp and Kc for the reaction? NH4HS (s) NH3 (g) + H2S (g)

Fourth lecture Reaction Quotient Qc The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression. Qc > Kc system proceeds to left to reach equilibrium Qc = Kc the system is at equilibrium Qc < Kc system proceeds to right to reach equilibrium

𝐐𝐜= 𝐍𝐚𝐂𝐥 𝐇𝐂𝐥 [𝐍𝐚𝐎𝐇] = 𝟔 (𝟑.𝟐) (𝟒.𝟑) = 0.436
Fourth lecture Find the value of Q and determine which side of the reaction is favored. Given Keq=0.5 HCl (g) + NaOH (aq) ⇌ NaCl (aq)+ H2O (l) [HCl]= 3.2 M [NaOH]= 4.3 M [NaCl]=6 M 𝐐𝐜= 𝐍𝐚𝐂𝐥 𝐇𝐂𝐥 [𝐍𝐚𝐎𝐇] = 𝟔 (𝟑.𝟐) (𝟒.𝟑) = 0.436 Qc = … Q is less than Keq so the reaction shifts RIGHT, favors the products.

Chemical Kinetics & Chemical Equilibrium
Fourth lecture Chemical Kinetics & Chemical Equilibrium ratef = kf [A][B]2 A + 2B AB2 kf kr rater = kr [AB2] Equilibrium ratef = rater kf [A][B]2 = kr [AB2] kf kr [AB2] [A][B]2 = Kc =

Equilibrium Constant Calculations
Fourth lecture Equilibrium Constant Calculations If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. A + B C + D Kc ′ Kc = ′ [C][D] [A][B] Kc = ′ [E][F] [C][D] C + D E + F Kc ′′ Kc [E][F] [A][B] Kc = A + B E + F Kc = Kc ′′ ′ x

Equilibrium Constant Calculations
Fourth lecture Equilibrium Constant Calculations When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. N2O4 (g) NO2 (g) 2NO2 (g) N2O4 (g) = 4.63 x 10-3 K = [NO2]2 [N2O4] K = [N2O4] [NO2]2 ′ = 1 K = 216

Equilibrium Calculations
Fourth lecture Equilibrium Calculations A closed system initially containing 0.001M H2 and M I2 is allowed to reach equilibrium at 448oC. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3 M. Calculate Kc at 448oC for the reaction taking place, which is: What Do We Know? [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 Change At equilibrium 1.87 x 10-3

[HI] Increases by 1.87 x 10-3 M [H2], M [I2], M [HI], M Initially
Fourth lecture [HI] Increases by 1.87 x 10-3 M [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 Change +1.87 x 10-3 At equilibrium 1.87 x 10-3

Stoichiometry tells us [H2] and [I2] decrease by half as much …
Fourth lecture Stoichiometry tells us [H2] and [I2] decrease by half as much … [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 Change -9.35 x 10-4 +1.87 x 10-3 At equilibrium 1.87 x 10-3

Fourth lecture We can now calculate the equilibrium concentrations of all three compounds… [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 Change -9.35 x 10-4 +1.87 x 10-3 At equilibrium 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3

…and, therefore, the equilibrium constant is calculated as:
Fourth lecture …and, therefore, the equilibrium constant is calculated as:

Le Châtelier’s Principle
Fourth lecture Le Châtelier’s Principle If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Changes in Concentration N2 (g) + 3H2 (g) NH3 (g) Add NH3 Equilibrium shifts left to offset stress

Changes in Concentration continued
Fourth lecture Changes in Concentration continued Add Add aA + bB cC + dD Remove Remove Change Shifts the Equilibrium Increase concentration of product(s) left Decrease concentration of product(s) right Increase concentration of reactant(s) right Decrease concentration of reactant(s) left 92

II. Changes in Volume and Pressure
Fourth lecture II. Changes in Volume and Pressure A (g) + B (g) C (g) Change Shifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas Increase volume Side with most moles of gas Decrease volume Side with fewest moles of gas

III. Temperature Changes
Fourth lecture III. Temperature Changes Consider heat as a product in exothermic reactions Add heat  Shift to reactants Remove heat  Shift to products A + B = AB + Heat Consider heat as a reactant in endothermic reactions Add heat  Shift to products Remove heat  Shift to reactants A + B + heat = AB

Catalyst lowers Ea for both forward and reverse reactions.
Fourth lecture Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner Catalyst lowers Ea for both forward and reverse reactions.

Le Châtelier’s Principle - Summary
Fourth lecture Le Châtelier’s Principle - Summary Change Shift Equilibrium Change Equilibrium Constant Concentration yes no Pressure yes* no Volume yes* no Temperature yes yes Catalyst no no *Dependent on relative moles of gaseous reactants and products

Questions Question 1 Question 3 Question 4 Question 2 Fourth lecture
Which equilibrium in gaseous phase would be unaffected by an increase in pressure: (a) N2O4 ->2NO2 (b) N2 + O2 ->2NO (c) N2 + 3H2 ->2NH3 (d) CO + ½ O2 ->O2 + CO2 Question 2 For the equilibrium , 2NO2(g) -> N2O4(g) kcal An increase of temperature will: (a) Favour the formation of N2O4 (b) Favour the decomposition of N2O4 (c) Not affect the equilibrium (d) Stop the reaction Question 3 The equilibrium constant (Kc) for the reaction is 2SO3(g) -> 2SO2(g) + O2(g) system as described by the above equation is: (a) [SO2]2/[SO3] (b) [SO2]2[O2]/[SO3]2 (c) [SO3]2/[SO3]2[O2] (d) [SO2][O2] Question 4 At equilibrium, __________. (a) the rates of the forward and reverse reactions are equal (b) the rate constants of the forward and reverse reactions are equal (c) all chemical reactions have ceased (d) the value of the equilibrium constant is 1

Questions Question 5 Question 7 Question 8 Question 6
Fourth lecture Question 5 The value of Keq for the following reaction is 0.25: SO2 (g) + NO2 (g) -> SO3 (g) + NO (g) The value of Keq at the same temperature for the reaction below is __________. 2SO2 (g) + 2NO2 (g) -> 2SO3 (g) + 2NO (g) 0.062 16 0.25 0.50 Question 6 Consider the reaction: 2 SO2(g) + O2(g) ↔ 2 SO3(g). If, at equilibrium at a certain temperature, [SO2] = 1.50 M, [O2] = M, and [SO3] = 1.25 M, what is the value of the equilibrium constant? (a) (b) 6.94 (c) (d) 0.14 Question 7 What is the correct equilibrium constant expression for the following reaction? 2 Cu(s) + O2(g) → 2 CuO(s) (a) Keq = 1/[O2]2 (b) Keq = [CuO]2/[Cu]2[O2] (c) Keq = [O2] (d) Keq = 1/[O2] Question 8 What is the relationship of the equilibrium constants for the following two reactions? (1) 2 NO2(g) ↔ N2O4(g); K1 (2) N2O4(g) ↔ 2 NO2(g) K2 (a) K1= 1/ K2 (b) K2= 1/ K1 (c) K1= K2 (d) both a and b are correct

Questions Question 9 Question 11 Question 12 Question 10
Fourth lecture Question 9 Consider the following endothermic reaction: H2(g) + I2(g) ↔ 2 HI(g). If the temperature is increased, (a) more HI will be produced (b) some HI will decompose, forming H2 and I2 (c) the magnitude of the equilibrium constant will decrease (d) the pressure in the container will increase Question 10 Consider the following reaction at equilibrium: NO2(g) + CO(g) ↔ NO(g) + CO2(g). Suppose the volume of the system is decreased at constant temperature, what change will this cause in the system? (a) A shift to produce more NO (b) A shift to produce more CO (c) A shift to produce more NO2 (d) No shift will occur Question 11 Which of these four factors can change the value of the equilibrium constant? (a) catalyst (b) pressure (c) concentration (d) temperature Question 12 Which general rule helps predict the shift in direction of an equilibrium reaction? (a) Le Chatelier's principle (b) Haber process (c) Equilibrium constant (d) Bosch theory

Type of Chemical Reactions in Aqueous Solutions
Fifth lecture Type of Chemical Reactions in Aqueous Solutions Ionic equilibrium Chapter 5

Type of Chemical Reactions in Aqueous Solutions
Fifth lecture Type of Chemical Reactions in Aqueous Solutions Acid-Base Reactions Oxidation-Reduction Reactions Precipitation Reactions

acid + base → salt + water HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
Fifth lecture I. Acid-Base Reactions acid + base → salt + water HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) 102

II. Oxidation-Reduction Reactions
Fifth lecture II. Oxidation-Reduction Reactions Redox reactions are electron transfer reactions Mg + 2Ag+ →Mg2++2Ag Half-reactions: Mg (s) → Mg2++ 2e 2Ag+ + 2e→2Ag Oxidation Reactions : half-reaction that involves a loss of electrons Reduction Reactions : half-reaction that involves a gain of electrons

III. Precipitation Reactions
Fifth lecture III. Precipitation Reactions A precipitate is an insoluble solid that separates from the solutions Pb(NO3)2 (aq) + 2KI (aq) → PbI2 (s) + 2KNO3 (aq) Pb2+ (aq) +2NO3- (aq) + 2K+ (aq) + 2I- (aq) → PbI2 (s) + 2K+ (aq) +2NO3- (aq) Pb2+ (aq) + 2I- (aq) → PbI2 (s)

Definition of acids and bases
Fifth lecture Acids & Bases Definition of acids and bases Arrhenius concept Brønsted-Lowry concept Lewis concept 105

1- Arrhenius Concept HCl (aq) H+ (aq) + Cl- (aq)
Fifth lecture 1- Arrhenius Concept An acid is a compound that releases H+ ions in water A base is a compound that releases OH- in water. HCl (aq) H+ (aq) + Cl- (aq) NaOH (aq) Na+ (aq) + OH- (aq) Limitations: Some bases do not contain OH- 106

2- Brønsted-Lowry Concept
Fifth lecture 2- Brønsted-Lowry Concept An acid is any molecule or ion that can donate a proton H+. A base is any molecule or ion can accept a proton. proton-transfer reaction 107

Fifth lecture 3- Lewis Concept An acid as an electron pair acceptor and a base as an electron pair donor. + N H : B F Another examples: hydration of AlCl3, BCl3, OH- 108

Strength of Acids and Bases
Fifth lecture Strength of Acids and Bases A strong acid or base ionizes completely in water Strong Acids Strong bases HCl LiOH HBr NaOH HI KOH HNO3 Ca(OH) 2 H2SO4 Sr(OH) 2 HClO4 Ba(OH) 2 109

A weak acid or base ionizes only to a limited extent in water
Fifth lecture Weak Acids and Bases A weak acid or base ionizes only to a limited extent in water Examples: CH3COOH, NH3

Acid or Base Ionization Constant
Fifth lecture Acid or Base Ionization Constant It is a measure of the strength of acid or base. The ionization constant has the same equilibrium expression. CH3COOH + H2O CH3COO- + H3O+ NH3 + H2O NH4+ + HO- 111

Self-ionization of water
Fifth lecture Self-ionization of water Water acts either as an acid or a base Or Kw = water dissociation constant 112

Self-ionization of water
Fifth lecture Self-ionization of water At 25°C, you observe the following conditions. an acidic solution, [H+] > [OH-] a neutral solution, [H+] = [OH-] a basic solution, [H+] < [OH-] 113

Fifth lecture pH of Solutions The pH of a solution is defined as the negative logarithm of the molar hydrogen-ion concentration In a neutral solution, whose hydrogen-ion concentration is 1.0 x 10-7, the pH = 7.00 114

In an acidic solution, [H+] > 1.0 x 10-7 M, pH<7
Fifth lecture At 25°C, you observe the following conditions In an acidic solution, [H+] > 1.0 x 10-7 M, pH<7 In a neutral solution, [H+] = 1.0 x 10-7 M, pH=7 In a basic solution, [H+] < 1.0 x 10-7 M, pH>7 115

Fifth lecture Example For a solution in which the hydrogen-ion concentration is 1.0 x 10-3, the pH is: Note that the number of decimal places in the pH equals the number of significant figures in the hydrogen-ion concentration 116

Fifth lecture Examples The hydrogen ion concentration of a fruit juice is 3.3 x 10-2 M. What is the pH of the juice? Is it acidic or basic? If a solution has pH of 5.50, calculate its [OH-] 117

pH of Strong Acids and Bases
Fifth lecture pH of Strong Acids and Bases Dissociation of a strong base: NaOH Na+ + OH- complete dissociation of a base and no base in the form of NaOH will be left in solution pOH = -log[OH-] pH = 14 - pOH = 14 + log [OH-]

Fifth lecture Example An ammonia solution has a hydroxide-ion concentration of 1.9 x 10-3 M. What is the pH of the solution? You first calculate the pOH: Then the pH is: 119

pH of Weak Acids and Bases
Fifth lecture pH of Weak Acids and Bases Examples: Ka (HF)=7.1 x , Ka (HCOOH)=1.7 x 10-4 HA A- + H Ka = = = [A-][H+] x2 x2 c [HA] c-x c-x x x c-x = concentration of an acid at equilibrium c >> x for diluted weak acids x = concentration of products at equilibrium c = concentration of an acid at the beginning [H+] = x = (Ka c)1/2 pH = -log [H+] = -log (Ka c)1/2 pKa = -logKa

Questions Question 1 Question 4 Question 2 Question 5 Question 3
Fifth lecture Question 1 The solution with the lowest pH is A. 1.0M HF B. 1.0M HCN C. 1.0M HCOOH D. 1.0M CH3COOH Question 2 As the [H3O+] in a solution decreases, the [OH-] A. increases and the pH increases B. increases and the pH decreases C. decreases and the pH increases D. decreases and the pH decreases Question 3 The value of pKw at 25°C is A. 1.0 x 10-14 B. 1.0 x 10-7 C. 7.00 D Question 4 Which of the following describes the relationship between [H3O+] and [OH-]? A. [H3O+][OH-] = 14.00 B. [H3O+] + [OH-] = 14.00 C. [H3O+][OH-] = 1.0 x 10-14 D. [H3O+] + [OH-] = 1.0 x 10-14 Question 5 What is the pOH of 0.1 M NaOH? A. 1 B C. 0.40 D Question 6 The pH of a solution for which [OH–] = 1.0 x 10–6 is 1.00 8.00 6.00 –6.00 121

Questions Question 7 Question 10 Question 11 Question 8 Question 9
Fifth lecture Question 7 The ionization of water at room temperature is represented by A. H2O = 2H+ + O2- B. 2H2O = 2H2 + O2 C. 2H2O = H2 + 2OH- D. 2H2O = H3O+ + OH- Question 8 According to the Bronsted-Lowry theory, a base is a(n) proton donor proton acceptor electron donor electron acceptor Question 9 the pH of 1.0 M acetic acid (Ka is 1.86 x10-5 at 20 °C). 1.37 2.37 3.73 4.73 Question 10 Addition of HCl to water causes A. both [H3O+] and [OH-] to increase B. both [H3O+] and [OH-] to decrease C. [H3O+] to increase and [OH-] to decrease D. [H3O+] to decrease and [OH-] to increase Question 11 Which of the following statements concerning Arrhenius acids and Arrhenius bases is correct? In the pure state, Arrhenius acids are covalent compounds. In the pure state, Arrhenius bases are ionic compounds Dissociation is the process by which Arrhenius acids produce H+ ions in solution Arrhenius bases are also called hydroxide bases 122

Sixth lecture Chapter 6 Thermochemistry

Energy Energy is the capacity to do work.
Sixth lecture Energy Energy is the capacity to do work. Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Potential energy is the energy available by virtue of an object’s position

Kinds of Systems Closed system Open system Isolated system
Sixth lecture Kinds of Systems Open system can exchange mass and energy Closed system allows the transfer of energy (heat) but not mass Isolated system doesn't allow transfer of either mass or energy

Sixth lecture Examples

Sixth lecture Thermodynamics Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy

Sixth lecture Heat (q) Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the thermal energy Temperature = Thermal Energy

First Law: Energy of the Universe is Constant E = q + w
Sixth lecture First Law of Thermodynamics First Law: Energy of the Universe is Constant E = q + w q = heat. Transferred between two bodies w = work. Force acting over a distance (F x d) w = F x d

Thermodynamic State Functions
Sixth lecture Thermodynamic State Functions Thermodynamic State Functions: Thermodynamic properties that are dependent on the state of the system only regardless of the pathway. Examples: (Energy, pressure, volume, temperature) Other variables will be dependent on pathway (Examples: q and w). These are Path Functions. The pathway from one state to the other must be defined. DE = Efinal - Einitial DP = Pfinal - Pinitial DV = Vfinal - Vinitial DT = Tfinal - Tinitial 130

Sixth lecture Thermochemistry Thermochemistry is the study of heat change in chemical reactions. Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2H2 (g) + O2 (g) H2O (l) + energy H2O (g) H2O (l) + energy Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + 2HgO (s) Hg (l) + O2 (g) energy + H2O (s) H2O (l) 131

Enthalpy of Chemical Reactions
Sixth lecture Enthalpy of Chemical Reactions Definition of Enthalpy Thermodynamic Definition of Enthalpy (H): H = E + PV E = energy of the system P = pressure of the system V = volume of the system

Changes in Enthalpy (ΔH)
Sixth lecture Changes in Enthalpy (ΔH) Consider the following expression for a chemical process: DH = Hproducts - Hreactants If DH >0, then qp >0. (+) The reaction is endothermic If DH <0, then qp <0. (-) The reaction is exothermic DH=qp qp : heat at constant pressure Calorimetry: the measurement of heat change 133

(chemical reactions or physical changes) Endothermic processes
Sixth lecture Kinds of Processes (chemical reactions or physical changes) Endothermic processes Exothermic processes H2O (s) CH4 (g) + 2O2 (g) Heat released from system to surroundings Heat absorbed from system to surroundings q = + q = - H2O (l) CO2 (g) + 2H2O (l) 12

Heat Capacity C = m s q = C ΔT Heat Capacity (C) Specific heat (S)
Sixth lecture Heat Capacity Heat Capacity (C) Specific heat (S) The amount of heat required to raise the temperature of a given quantity of the substance by one degree Celsius or kelvin. The amount of heat required to raise the temperature of one gram of the substance by one degree Celsius or kelvin. C = m s J/oC J/K J/g.oC J/g.K mass q = C ΔT

Specific heats of some common substances
Sixth lecture Specific heats of some common substances Specific Heat (J/g.oC) Substance 0.900 Al 0.129 Au 0.720 C (graphite) 0.502 C (diamond) 0.385 Cu 0.444 Fe 0.139 Hg 4.184 H2O 2.46 C2H5OH 15

Standard Enthalpy (Heat) of reaction (ΔHorxn)
Sixth lecture Standard Enthalpy (Heat) of reaction (ΔHorxn) Enthalpy change at standard conditions (25 oC, 1 atm) N2 (g) + 3H2 (g) NH3 (g) ΔHo = kJ Thermochemical reaction

Standard Heat of formation (ΔHfo)
Sixth lecture Standard Heat of formation (ΔHfo) The heat change that results when 1 mol of the compound is formed from standard state of its elements The standard enthalpy of formation of any element in its most stable form is zero. DH0 (C, diamond) = 1.90 kJ/mol What is ΔHfo of O2 (g), Hg(l), C(graphite)?

Thermochemical Equations
Sixth lecture Thermochemical Equations CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = kJ/mol It shows the physical states of all products and reactants Balanced It shows Heat of reaction kJ H2O (s) H2O (l) DH = 6.01 kJ/mol If you reverse a reaction, the sign of DH changes H2O (l) H2O (s) DH = kJ/mol If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2H2O (s) H2O (l) DH = 2 x 6.01 = 12.0 kJ

ΔHo = ∑ n ΔHfo (products) - ∑ n ΔHfo (reactants)
Sixth lecture How to calculate ΔHro 1- Direct method indirect method 1- Direct method: by standard heat of formation ΔHo = ∑ n ΔHfo (products) - ∑ n ΔHfo (reactants) n = no. of moles in the balanced thermochemical equation

Example 2Al (s) + Fe2O3 (s) Al2O3 (s) + 2Fe(l)
Sixth lecture Example Calculate the enthalpy of the following reaction: 2Al (s) + Fe2O3 (s) Al2O3 (s) + 2Fe(l) ΔHfo of Fe2O3, Al2O3 and Fe(l) = , and kJ/mol ΔHo = ∑ n ΔHfo (products) - ∑ n ΔHfo (reactants) ΔHo = [(ΔHfo (Al2O3))+ (2× ΔHfo (Fe))]-[(2×ΔHfo (Al))+(ΔHfo (Fe2O3))] ΔHo = [ ( )+ (2×12.40)] – [2×(0)+(-822.2)] = kJ

2- indirect method :(Hess’s Law)
Sixth lecture 2- indirect method :(Hess’s Law) DH for a process involving the transformation of reactants into products is not dependent on pathway. Therefore, we can pick any pathway to calculate DH for a reaction. When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. DH is a state function

2NO2(g) N2(g) + 2O2(g) DH = -68 kJ
Sixth lecture Hess’ Law Details Once can always reverse the direction of a reaction when making a combined reaction. When you do this, the sign of DH changes. N2(g) + 2O2(g) NO2(g) DH = 68 kJ 2NO2(g) N2(g) + 2O2(g) DH = -68 kJ

N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ 2N2(g) + 4O2(g) 4NO2(g) DH = 136 kJ
Sixth lecture If the coefficients of a reaction are multiplied by a constant, the value of DH is also multiplied by the same integer. N2(g) + 2O2(g) NO2(g) DH = 68 kJ 2N2(g) + 4O2(g) NO2(g) DH = 136 kJ

Example Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g)
Sixth lecture Example Calculate the enthalpy of the following reaction: Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) ΔHfo of Fe2O3, CO and CO2= , and – kJ/mol 1- direct method ΔHo = ∑ n ΔHfo (products) - ∑ n ΔHfo (reactants) ΔHo = [(2× ΔHfo (Fe))+ (3× ΔHfo (CO2))]- [(ΔHfo (Fe2O3)+(3xΔHfo (CO))] ΔHo = [ (0)+ (3×(-393.5))] – [(-822.2)+(3×(-110.5))] ΔHo = kJ

Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g)
Sixth lecture 2- Hess’s Law Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) 1- CO (g) + ½O2 (g) CO2 (g) ΔHo = kJ 2- 2Fe (s) + 3/2O2 (g) Fe2O3 (s) ΔHo = kJ 3CO (g) + 3/2O2 (g) CO2 (g) ΔHo = 3× = kJ Fe2O3 (s) Fe (s) + 3/2O2 (g) ΔHo = kJ Fe2O3 (s) + 3CO (g) Fe (s) + 3CO2 (g) ΔHo = kJ 23

Questions Question 1 Question 3 Question 2 Question 4 Sixth lecture
An exothermic reaction causes the surroundings to: become basic decrease in temperature condense increase in temperature decrease in pressure Question 2 How much heat is evolved when 320 g of SO2 is burned according to the chemical equation shown below? 2 SO2(g) + O2(g) ----> 2 SO3(g) ΔHorxn = -198 kJ 5.04 x 10-2 kJ 9.9 x 102 kJ 207 kJ 5.0 x 102 kJ None of the above Question 3 The specific heat of aluminum is cal/g.oC. Determine the energy, in calories, necessary to raise the temperature of a 55.5 g piece of aluminum from 23.0 to 48.6oC. 109 cal 273 cal 577 cal 347 cal 304 cal Question 4 Energy is the ability to do work and can be: converted to one form to another can be created and destroyed used within a system without consequences none of the above 147

Questions Question 5 Question 7 Question 6 Question 8 Sixth lecture
To which one of the following reactions, occurring at 25oC, does the symbol ΔHof [H2SO4(l)] refer? H2(g) + S(s) + 2 O2(g) ----> H2SO4(l) H2SO4(l) ----> H2(g) + S(s) + 2 O2(g) H2(g) + S(g) + 2 O2(g) ----> H2SO4(l) H2SO4(l) ----> 2 H(g) + S(s) + 4 O(g) 2 H(g) + S(g) + 4 O(g) ----> H2SO4(l) Question 6 Given: SO2(g) + ½O2(g) ----> SO3(g) ΔHorxn = -99 kJ, what is the enthalpy change for the following reaction? 2 SO3(g) ----> O2(g) + 2 SO2(g) 99 kJ -99 kJ 49.5 kJ -198 kJ 198 kJ Question 7 The specific heat of aluminum is cal/g.oC. Determine the energy, in calories, necessary to raise the temperature of a 55.5 g piece of aluminum from 23.0 to 48.6oC. 109 cal 273 cal 577 cal 347 cal 304 cal Question 8 Standard enthalpy of reactions can be calculated from standard enthalpies of formation of reactants. True False 148

Questions Question 9 Question 10 Sixth lecture
Calculate ΔHorxn for the combustion reaction of CH4 shown below given the following: ΔHof CH4(g) = kJ/mol; ΔHof CO2(g) = kJ/mol; ΔHof H2O(l) = kJ/mol. CH4(g) + 2 O2(g) ----> CO2(g) + 2 H2O(l) kJ 889.7 kJ kJ kJ None of the above Question 10 Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) ----> 2 CO2(g) + 2 H2O(l) ΔHorxn = kJ; C(s) + O2(g) ----> CO2(g) ΔHof = kJ; H2(g) + ½O2(g) ----> H2O(l) ΔHof = kJ 731 kJ 2.77 x 103 kJ 1.41 x 103 kJ 87 kJ 52 kJ 149

Seventh lecture Chapter 7 Gases

Elements that exist as gases at 250C and 1 atmosphere
Seventh lecture Elements that exist as gases at 250C and 1 atmosphere

Seventh lecture

Physical Characteristics of Gases
Seventh lecture Physical Characteristics of Gases Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids.

Force Pressure = Area Units of Pressure 1 pascal (Pa) = 1 N/m2
Seventh lecture Force Area Pressure = (force = mass x acceleration) Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa

Boyle’s law: The volume of a given amount of gas held at
constant temperature varies inversely with the applied pressure. Seventh lecture 1 atm 2 atm 2 Liters 4 Liters P a 1/V T = Constant n = Constant P x V = constant P1 x V1 = P2 x V2

P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL
Seventh lecture A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 V2 726 mmHg x 946 mL 154 mL = P2 = = 4460 mmHg

As T increases V increases
Seventh lecture Variation in Gas Volume with Temperature at Constant Pressure As T increases V increases

Seventh lecture Charles’s Law: The volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature . V a T V = constant x T V1/T1 = V2 /T2 Temperature must be in Kelvin T (K) = t (0C)

Seventh lecture A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1 /T1 = V2 /T2 V1 = 3.20 L V2 = 1.54 L T1 = K T2 = ? T1 = 125 (0C) (K) = K V2 x T1 V1 1.54 L x K 3.20 L = T2 = = 192 K

Seventh lecture Gay-Lussac’s Law: The pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? P1 T1 P2 T2 = P1 = 1.20 atm T1 = 291 K P2 = ? T2 = 358 K P2 = P1 x T2 T1 = 1.20 atm x 358 K 291 K = 1.48 atm

Boyle’s P V T, n Charles’ V T P, n Gay-Lussac’s P T V, n
Seventh lecture Summary LAW RELAT-IONSHIP CON-STANT Boyle’s P V P1V1 = P2V2 T, n Charles’ V T V1/T1 = V2/T2 P, n Gay-Lussac’s P T P1/T1 = P2/T2 V, n

Avogadro’s Law V a number of moles (n) V = constant x n
Seventh lecture Avogadro’s Law Constant temperature Constant pressure V a number of moles (n) V = constant x n V1 / n1 = V2 / n2

4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P
Seventh lecture Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH3 + 5O NO + 6H2O 1 mole NH mole NO At constant T and P 1 volume NH volume NO

Ideal Gas Equation 1 Boyle’s law: P a (at constant n and T) V
Seventh lecture Ideal Gas Equation Boyle’s law: P a (at constant n and T) 1 V Charles’ law: V a T (at constant n and P) Avogadro’s law: V a n (at constant P and T) V a nT P V = constant x = R nT P R is the gas constant PV = nRT

PV = nRT PV (1 atm)(22.414L) R = = nT (1 mol)(273.15 K)
Seventh lecture The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies L. PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)( K) R = L • atm / (mol • K)

PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.7 L
Seventh lecture What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = K P = 1 atm PV = nRT n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = nRT P V = 1 atm 1.37 mol x x K L•atm mol•K V = 30.7 L

Density (d) Calculations
Seventh lecture Density (d) Calculations m is the mass of the gas in g d = m V = PM RT M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L

dRT P M = d = m V = = 2.21 1 atm x 0.0821 x 300.15 K M = M =
Seventh lecture A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas? dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L 2.21 g L 1 atm x x K L•atm mol•K M = M = 54.5 g/mol

Seventh lecture Dalton’s law of partial pressures: the total pressure, Ptotal, of a mixture of gases is the sum of their individual gas partial pressures V and T are constant Ptotal = P1+ P2 P11 P2 169

PA = nART V PB = nBRT V XA = nA nA + nB XB = nB nA + nB PT = PA + PB
Seventh lecture Consider a case in which two gases, A and B, are in a container of volume V. PA = nART V nA is the number of moles of A PB = nBRT V nB is the number of moles of B XA = nA nA + nB XB = nB nA + nB PT = PA + PB PA = XA PT PB = XB PT Pi = Xi PT mole fraction (Xi ) = ni nT

Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane =
Seventh lecture A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = Xi PT PT = 1.37 atm 0.116 Xpropane = = Ppropane = x 1.37 atm = atm

Kinetic Molecular Theory of Gases
Seventh lecture A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic. Gas molecules exert neither attractive nor repulsive forces on one another. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy KE = ½ mu2

Compressibility of Gases
Seventh lecture Compressibility of Gases Boyle’s Law P a collision rate with wall Collision rate a number density Number density a 1/V P a 1/V Gay-Lussac’s Law Collision rate a average kinetic energy of gas molecules Average kinetic energy a T P a T

Dalton’s Law of Partial Pressures
Seventh lecture Avogadro’s Law P a collision rate with wall Collision rate a number density Number density a n P a n Dalton’s Law of Partial Pressures Molecules do not attract or repel one another P exerted by one type of molecule is unaffected by the presence of another gas Ptotal = SPi

Seventh lecture Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. NH4Cl NH3 17 g/mol HCl 36 g/mol

Seventh lecture Gas effusion is the is the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening. = r1 r2 t2 t1 M2 M1  Nickel forms a gaseous compound of the formula Ni(CO)x What is the value of x given that under the same conditions methane (CH4) effuses 3.3 times faster than the compound? M2 = r1 r2 ( ) 2 x M1 r1 = 3.3 x r2 = (3.3)2 x 16 = 174.2 M1 = 16 g/mol x • 28 = 174.2 x = 4.1 ~ 4

Deviations from Ideal Behavior
Seventh lecture Deviations from Ideal Behavior 1 mole of ideal gas PV = nRT n = PV RT = 1.0

( ) } } Van der Waals equation Non ideal gas an2 P + (V – nb) = nRT V2
Seventh lecture Van der Waals equation Non ideal gas ( an2 V2 ) P (V – nb) = nRT } corrected pressure } corrected volume

Questions Seventh lecture
Which of the following is not a characteristic of substances in the gas phase? A)Substances in the gas phase have much lower densities than the same substances would have in the liquid or solid phase. B)A mixture of substances in the gas phase will form a homogeneous solution, whereas the same mixture might not form a homogeneous solution in the liquid phase. C)Substances in the gas phase retain their shapes easily. D)Substances in the gas phase are compressible. 2. A sample of gas occupies 2.78 x 103 mL at 25oC and 760 mm Hg. What volume will the gas sample occupy at the same temperature and 475 mm Hg? A) L B) 1.04 L C) 1.74 L D) 4.45 L E) None of the above 3.A steel tank contains carbon dioxide at a pressure of 13.0 atm when the temperature is 34oC. What will be the internal gas pressure when the tank and its contents are heated to 100oC. A) 38.2 atm B) 9.40 atm C) 10.7 atm D) 15.8 atm E) None of the above. 4.Calculate the density of nitrogen gas, in grams per liter, at STP. A) g/L B) g/L C) 1.25 g/L D) 2.50 g/L E) None of the above

5. A gas evolved during the fermentation of alcohol had a volume of 19.4 L at 17oC and 746 mm Hg. How many moles of gas were collected? A) 1.25 mol B) mol C) 10.5 mol D) 13.6 mol E) 608 mol 6. How many grams of carbon dioxide are contained in 550 mL of this gas at STP? A) g B) g C) 1080 g D) g E) 1.1 g 7. A g sample of an unknown vapor occupies 368 mL at 114oC and 946 mm Hg. The empirical formula of the compound is NO2. What is the molecular formula of the compound? A) NO2 B) N4O8 C) N3O6 D) N2O4 E) N5O10. 8.An organic compound was analyzed and found to contain 55.8% C, 7.03% H, and 37.2% O. A g sample of the compound was vaporized and found to occupy 530 cm3 at 100oC and 740 torr. Which of the following is the correct molecular formula of the compound? A) C2H3O B) C6H4O2 C) C3H2O D) C4H6O2 E) C2H3O2

9.What volume of chlorine gas at 646 torr and 32oC would be produced by the reaction of g of MnO2 according to the following chemical equation? MnO2(s) + 4 HCl(aq) ----> MnCl2(aq) + Cl2(g) + 2 H2O(l) A) 5.00 L B) L C) 2.33 L D) L E) None of the above 10. A mixture of neon, argon, and xenon had a total pressure of 1560 mm Hg at 298 K. The mixture was found to contain 1.50 mol Ne, 2.65 mol Ar, and mol Xe. What is the partial pressure of Xe? A) 701 mm Hg B) 658 mm Hg C) 396 mm Hg D) 463 mm Hg 11.A steel tank contains carbon dioxide at a pressure of 13.0 atm when the temperature is 34oC. What will be the internal gas pressure when the tank and its contents are heated to 100oC. A) 38.2 atm B) 9.40 atm C) 10.7 atm D) 15.8 atm E) None of the above. 12.Which of the following correctly identifies Boyle's law? A) PV=k1 B) V=k2T 13.The magnitude of one Kelvin, one Celsius degree, and one degree on the absolute temperature scale is the same. A) True B) False

14.The Kelvin temperature scale is useful when comparing: various gas samples at different densities volume of a gas sample with temperature at constant pressure C) pressure of gas samples at different volumes and constant temperature D) various liquids at constant pressure 15. Which of the following is not an assumption of the kinetic theory of gases? Elasticity refers to the molecules random interactions resulting in no net energy change. B) Gas molecules are viewed as points due to large distances between them. C) Gas molecules can attract other gas molecules. D) Temperature in Kelvin and average kinetic energy are proportional. 16. A sample of CO2(g) has a volume of 2L at pressure P and temperature T. If the pressure becomes triple the original value, at the same absolute temperature, the volume of CO2 will be A) L B) 2/3 L C) 6L D) 2L 17. Using the van der Waals equation, calculate the pressure exerted by 15.0 mol of carbon dioxide confined to a 3.00-L vessel at 56.0 oC. Note: Values for a and b in the van der Waals equation: a = L2.atm/mol, and b = L/mol. A) 23.2 atm B) 2.16 atm C) 81.9 atm D) 81.9 atm

Eighth lecture Chapter 8 Electrochemistry

Oxidation half-reaction (lose e-)
Eighth lecture Electrochemical processes are oxidation-reduction reactions in which: the energy released by a spontaneous reaction is converted to electricity or electrical energy is used to cause a nonspontaneous reaction to occur 2+ 2- 2Mg (s) + O2 (g) MgO (s) 2Mg Mg2+ + 4e- Oxidation half-reaction (lose e-) O2 + 4e O2- Reduction half-reaction (gain e-)

Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)
Eighth lecture Oxidation-Reduction Reactions Oxidizing Agent- a substance that accepts electrons from another substance, causing the other substance to be oxidized. Reducing Agent- a substance that donates electrons to another substance, causing it to become reduced. Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s) Zn Zn2+ + 2e- Zn is oxidized Zn is the reducing agent Cu2+ + 2e Cu Cu2+ is reduced Cu2+ is the oxidizing agent

Galvanic Cells anode oxidation cathode reduction spontaneous
Eighth lecture anode oxidation cathode reduction spontaneous redox reaction

electromotive force (emf) cell potential
Eighth lecture The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq) [Cu2+] = 1 M and [Zn2+] = 1 M Cell Diagram phase boundary Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s) anode salt bridge cathode

Standard Reduction Potentials
Eighth lecture Standard Reduction Potentials Standard reduction potential (E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. Reduction Reaction 2e- + 2H+ (1 M) H2 (1 atm) E0 = 0 V Standard hydrogen electrode (SHE)

Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
Eighth lecture E0 = 0.76 V cell Standard emf (E0 ) cell E0 = Ecathode - Eanode cell E0 = EH /H - EZn /Zn cell + 2+ 2 0.76 V = 0 - EZn /Zn 2+ EZn /Zn = V 2+ Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s) Anode (oxidation): Zn (s) Zn2+ (1 M) + 2e- Cathode (reduction): 2e- + 2H+ (1 M) H2 (1 atm) Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)

Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)
Eighth lecture E0 = 0.34 V cell E0 = Ecathode - Eanode cell Ecell = ECu /Cu – EH /H 2+ + 2 0.34 = ECu /Cu - 0 2+ ECu /Cu = 0.34 V 2+ Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s) Anode (oxidation): H2 (1 atm) H+ (1 M) + 2e- Cathode (reduction): 2e- + Cu2+ (1 M) Cu (s) H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)

E0 is for the reaction as written
Eighth lecture E0 is for the reaction as written The more positive E0 the greater the tendency for the substance to be reduced The half-cell reactions are reversible The sign of E0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

Cd is the stronger oxidizer
Eighth lecture What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution? Cd2+ (aq) + 2e Cd (s) E0 = V Cd is the stronger oxidizer Cd will oxidize Cr Cr3+ (aq) + 3e Cr (s) E0 = V Anode (oxidation): Cr (s) Cr3+ (1 M) + 3e- x 2 Cathode (reduction): 2e- + Cd2+ (1 M) Cd (s) x 3 2Cr (s) + 3Cd2+ (1 M) Cd (s) + 2Cr3+ (1 M) E0 = Ecathode - Eanode cell E0 = – (-0.74) cell E0 = 0.34 V cell

Spontaneity of Redox Reactions
Eighth lecture Spontaneity of Redox Reactions DG0 = -nFEcell

The Effect of Concentration on Cell Emf
Eighth lecture The Effect of Concentration on Cell Emf DG = DG0 + RT ln Q DG = -nFE DG0 = -nFE -nFE = -nFE0 + RT ln Q Nernst equation E = E0 - ln Q RT nF At 298 K - V n ln Q E E = - V n log Q E E =

Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)
Eighth lecture Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.60 M and [Cd2+] = M? Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq) Oxidation: Cd Cd2+ + 2e- n = 2 Reduction: 2e- + Fe Fe E0 = EFe /Fe – ECd /Cd 2+ E0 = – (-0.40) - V n ln Q E E = E0 = V - V 2 ln -0.04 V E = 0.010 0.60 E = 0.013 E > 0 Spontaneous

Batteries Dry cell Leclanché cell Anode: Zn (s) Zn2+ (aq) + 2e-
Eighth lecture Batteries Dry cell Leclanché cell Anode: Zn (s) Zn2+ (aq) + 2e- Cathode: 2NH4 (aq) + 2MnO2 (s) + 2e Mn2O3 (s) + 2NH3 (aq) + H2O (l) + Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)

Lead storage battery Anode: Pb (s) + SO2- (aq) PbSO4 (s) + 2e-
Eighth lecture Lead storage battery Anode: Pb (s) + SO2- (aq) PbSO4 (s) + 2e- 4 Cathode: PbO2 (s) + 4H+ (aq) + SO2- (aq) + 2e PbSO4 (s) + 2H2O (l) 4 Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) PbSO4 (s) + 2H2O (l) 4

Eighth lecture A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning Anode: 2H2 (g) + 4OH- (aq) H2O (l) + 4e- Cathode: O2 (g) + 2H2O (l) + 4e OH- (aq) 2H2 (g) + O2 (g) H2O (l)

Eighth lecture Corrosion Corrosion is the term usually applied to the degradation of metals by an electrochemical process.

Cathodic Protection of an Iron Storage Tank
Eighth lecture Cathodic Protection of an Iron Storage Tank

Electrolysis of molten NaCl
Eighth lecture Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur. Electrolysis of Water Electrolysis of molten NaCl

Electrolysis and Mass Changes
Eighth lecture Electrolysis and Mass Changes charge (C) = current (A) x time (s) 1 mol e- = 96,500 C

Eighth lecture How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of A is passed through the cell for 1.5 hours? Anode: 2Cl- (l) Cl2 (g) + 2e- Cathode: 2 mole e- = 1 mole Ca Ca2+ (l) + 2e Ca (s) Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g) mol Ca = 0.452 C s x 1.5 hr x 3600 s hr 96,500 C 1 mol e- x 2 mol e- 1 mol Ca x = mol Ca = 0.50 g Ca

Questions Eighth lecture 1. Separating redox equations is simpler if the equation is separated into oxidation and reduction portions. (a) True (b) False 4. A galvanic cell is prepared using zinc and iron. Its cell notation is as follows: Zn|Zn2+(1 M)||Fe2+(1 M), Fe3+(1M)|Pt. Which of the following reactions occurs at the cathode? Fe3+ + e- ----> Fe2+ (b) Fe2+ ----> Fe3+ + e- (c) Zn ----> Zn2+ + 2 e- (d) Zn2+ + 2e- ----> Zn 2. Which of the following does not denote the difference in electrical potential between the anode and cathode? cell voltage (b) electromotive force (c) Voltmeter (d) cell potential 5. A cell can be prepared from zinc and iron. What is the Eocell for the cell that forms from the following half reactions? Fe3+ + e- ----> Fe2+, Eo = 0.77 V; Zn2+ + 2e- ----> Zn, Eo = V -1.53 V (b) V (c) 0.01V (d) 1.53 V 3. The same electrode may act as either the anode or the cathode depending on the second electrode used in setting up the galvanic cell (a) True (b) False 6. The electrode at which oxidation occurs is called the anode (b) cathode (c) salt bridge

Questions Eighth lecture 7. Calculate the free energy change per mole of Cu2+ formed in the following reaction at 25 oC. Cu + 2 Ag+ ----> 2 Ag + Cu2+. Cu2+ + 2 e- ----> Cu Eo = 0.34 V; Ag+ + e- ----> Ag, Eo = 0.80 V. 0.46 kJ (b) 89 kJ (c) 44.5 kJ (d) -89 kJ 9. Which is a secondary cell? Mercury cell. (b) dry Cell (c) H2-O2 Cell (d) Lead storage battery 10. The principal function of a fuel cell is to Produce fuel. (b) Electrolyze fuel. (c) Produce hydrogen. (d) Produce electricity 11. Corrosion is the disintegration of metals through combustion. (b) oxidation (c) reduction (d) exposure 8. One of the differences between a galvanic cell and an electrolytic cell is that in a galvanic cell an electric current is produced by a chemical reaction. electrons flow toward the anode. a non spontaneous reaction is forced to occur using an electric current from an external source. reduction occurs at the anode. 12. What conditions are required for rusting to take place? Water only (b) oxygen only (c) Water and oxygen (d) Water and carbon dioxide 13. What mass of copper can be deposited by the passage of 12.0 A for 25.0 min through a solution of copper(II) sulfate? [1 C = 1 A·s; F = 96,500 C] 5.93 g (b) 3.95 g (c) 1.97 g (d) g 205

General Chemistry 402101-4.

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