Thermochemistry Thermochemistry is the study of heat change in a chemical reaction. It is part of a broader topic called thermodynamics. This is the scientific study of the interconversion of heat and other kinds of energy.

Principles of Heat Flow Heat is a particular form of energy that is transferred from a body of high temperature to low temperature. Temperature is the measure of the average kinetic energy in particles. When heat flow is discussed we must differentiate between the system and it’s surroundings: The system is that part of the universe on which attention is focused. The surroundings, exchange energy with the system, make up the rest of the universe – really only include those materials in close contact with the system Energy cannot be created nor destroyed, simply converted or transferred – 1st Law of Thermodynamics, where ΔE = q + w

Systems Open system Closed system Isolated system Can exchange mass and energy, usually in the form heat, with its surroundings. Closed system Allows the transfer of energy (heat) but not mass. Isolated system Does not allow the transfer of either mass or energy

Heat (q) When heat flows from the system to its surroundings then the heat flow, symbolized by q, is negative. When heat flows into the system from its surroundings then the heat flow, q, is positive. An endothermic reaction: q > 0. Why? The decomposition of mercury II oxide: 2HgO (s)  2Hg (l) + O2(g) An exothermic reaction: q < 0. Why? The combustion of methane: CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l) We are not interested only in the direction of heat flow but its magnitude. Heat flow (q) is measured in the unit joules (J) and kilojoules (kJ). In the past chemists used to use the calorie as an energy unit. 1cal = 4.184 J 1 kcal = 4,184 J Be careful! – that calorie is not the same as the dieticians calorie – the dieticians calorie = 1 kcal.

Heat Flow & Heat Capacity The relationship between the magnitude of heat flow and the temperature change is given by the following equation: q = C x Δt q = heat flow, unit: J C = heat capacity represents the amount of heat required to raise the temp. of a system 1 *C, unit: J/*C Δt = final temperature – initial temperature (tf – ti), unit: *C

Heat Flow & Specific Heat Another relationship between the magnitude of heat flow and the temperature change is given by the following equation q = m x c x Δt m = mass, unit – grams c = specific heat represents the amount of heat required to raise 1 gram of a substance 1*C, unit – J/g *C specific heat, like density, is an intensive property and can be used to identify a substance. The specific heat of water = 4.18 J/g *C

Example 8.1 How much heat is given off when one mole of liquid water cools from 100.0 *C to 10.0 *C? Ans: -6.78 x 103 J

Coffee Cup Calorimetry Coffee cup calorimetry is used in general chemistry lab to determine the magnitude of heat (q). Consists of a polystyrene foam cup partially filled with a water. It has a tightly fitted lid in which an accurate thermometer is placed. Essentially all of the heat evolved by the reaction taking place within the calorimeter is absorbed by the water. So, the heat capacity of the coffee cup calorimeter is that of the water.

Calorimetry qreaction = -qcalorimeter qreaction = -Ccal x Δt qreaction = -mwater x cwater x Δt

qreaction = -qcalorimeter Bomb Calorimetry qreaction = -qcalorimeter qreaction = -Ccal x Δt

Example 8.2 When 1.00 gram of calcium chloride is added to 50.0 grams of water in a calorimeter, it dissolves and the temperature rises from 25.00 *C to 28.51 *C. Assuming that all the heat given off be the reaction is transferred to the water, calculate q for the reaction system. Ans: -734 J

Example 8.3 The reaction between hydrogen and chlorine can be studied in a bomb calorimeter. It is found that when a 1.00 g sample of hydrogen reacts completely, the temperature rises from 20.00 *C to 29.82 *C. Taking the heat capacity of the calorimeter to be 9.33 kJ/*C. Calculate the amount of heat evolved. Ans: -91.6 kJ

Enthalpy (H) Enthalpy is the measurement of heat (really energy) stored in chemical bonds. The heat flow for a reaction system is equal to the difference in enthalpy (H) between products and reactants keep in mind that this is at a constant pressure Enthalpy is a state function, depends only on the state of the system the way the system reached the state. ΔE and ΔH are both state functions unlike q and w. qreaction = ΔH = Hproducts - Hreactants

Exothermic vs Endothermic -Exothermic reactions: CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l) ΔH < 0 -Endothermic reactions: H2O (s)  H2O (l) ΔH > 0

Thermochemical Equations Equations that show the enthalpy relationship between products and reactants is called a thermochemical equation. For example: NH4NO3 (s)  NH4+(aq) + NO3-(aq) ΔH = +28.1 kJ Is the above reaction exothermic or endothermic? Why? H2 (g) + Cl2 (g)  2HCl (g) ΔH = -185 kJ

Thermochemical Equations Important Points The sign of ΔH indicates endothermic or exothermic reaction. The coefficients represent numbers of moles (i.e. ΔH = -185 kJ when there is 1 mole of hydrogen, 1 mole of chlorine, and 2 moles of hydrogen chloride). The phases of all species must be specified. Enthalpies will differ for the same substance in a different state of matter. The value given for ΔH applies when products and reactants are at the same temperature, generally 25 *C – unless otherwise stated.

Stoichiometry using Enthalpy The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by If aluminum is produced this way, how many grams of aluminum can form when 1.000 × 103 kJ of heat is transferred?

Steps to using enthalpy:

Example 8.4 Consider the thermochemical equation for the production of hydrogen chloride gas from it foundational elements. In this reaction ΔH = - 185 kJ. Calculate ΔH when: 1.00 mol of HCl is formed Ans: -92.5 kJ 1.00 g grams of chlorine reacts Ans: -2.61 kJ 2.50 L of HCl at 50.0 *C and 725 mmHg is formed. Ans: -8.32 kJ

Various Enthalpy Changes Some enthalpy changes are frequently studied and, therefore, they have special names: When 1 mol of a compound is produced from its elements, the heat of reaction is called the heat of formation (ΔHf): When 1 mol of a substance vaporizes, the enthalpy change is called the heat of vaporization (ΔHvap): When 1 mol of a substance melts, the enthalpy change is called the heat of fusion (ΔHfus):

Rules of Thermochemsitry 1. The magnitude of ΔH is directly proportional to the amount of reactant or product. For example, the more water you have the more heat it will take to boil it. 2. ΔH for a reaction is equal in magnitude but opposite in sign to ΔH for the reverse reaction. H2O (s)  H2O (l) ΔH = +6.00 kJ Based on the above reaction how much heat is evolved (or released) when liquid water freezes? 3. The value of ΔH for a reaction is the same whether it occurs in one step or in a series of steps. This is known as Hess’s law and is conveniently used for reactions that are difficult to carry out in a calorimeter: ΔH = ΔHequation 1 + ΔHequation 2 + …..

Example 8.5 Given the equation: 2H2 (g) + O2 (g)  2H2O (l) ΔH = -571.6 kJ Calculate the ΔH for the equation: H2O (l)  H2 (g) + ½ O2 (g) Ans: +285.8 kJ

Hess’s Law First, we identify Equation 3 as our “target” equation, the one whose ΔH we want to find, and we carefully note the number of moles of each reactant and product in it. We also note that ΔH1 and ΔH2 are the values for Equations 1 and 2 as written. Now we manipulate Equations 1 and/or 2 as follows to make them add up to Equation 3: •Equations 1 and 3 contain the same amount of S, so we leave Equation 1 unchanged. •Equation 2 has twice as much SO3 as Equation 3, so we multiply it by 1/2, being sure to halve ΔH2 as well. •With the targeted amounts of reactants and products now present, we add Equation 1 to the halved Equation 2 and cancel terms that appear on both sides:

To Summarize Hess’s Law Calculating an unknown ΔH basically involves three steps: Identify the target equation, the step whose ΔH is unknown, and note the number of moles of each reactant and product. 2. Manipulate the equations with known ΔH values so that the target numbers of moles of reactants and products are on the correct sides. Remember to: • Change the sign of ΔH when you reverse an equation. • Multiply numbers of moles and ΔH by the same factor. 3. Add the manipulated equations to obtain the target equation. All substances except those in the target equation must cancel. Add their ΔH values to obtain the unknown ΔH.

Example 8.6 Given: 1. C (s) + O2 (g)  CO2 (g) ΔH = -393.5 kJ 2. 2CO (g) + O2 (g)  2CO2 (g) ΔH = -566.0 kJ Calculate the enthalpy for: C (s) + ½ O2 (g)  CO (g) ΔH = ? Ans: -110.5 kJ

Finding ΔH from ΔHf Enthalpies of formation may be used to calculate the ΔH for the reaction. To do this apply this rule: The standard enthalpy change for a given thermochemical equation is equal to the sum of the standard enthalpies of formation of the product compounds minus the sum of the standard enthalpies of formation of the reactant compounds. ΔH = ∑ ΔHf products - ∑ ΔHf reactants

Using Heats of Formation When applying this equation, elements in their standard states can be omitted because their heats of formation equal zero. So for the following reaction: 2Al (s) + Fe2O3 (s)  2Fe (s) + Al2O3 (s) We can write: ΔH = ∑ ΔHf products - ∑ ΔHf reactants = ΔHf Al2O3 (s) - ΔHf Fe2O3 (s) The coefficients of the products and reactants must also be taken into account. So for the following reaction: 2Al (s) + 3Cu+2 (aq)  2Al+3 (aq) + 3Cu (s) ΔH = 2 ΔHf Al+3 (aq) - 3 ΔHf Cu+2 (aq)

Example 8.7 Using table 8.3 calculate (please refer to page 208 for balanced equations of the following reactions): ΔH for the combustion of methane -2219.9 kJ The ΔHf of benzene using the equation for the combustion of benzene when ΔH = -3267.4 kJ 49.0 kJ/mol The ΔH for the combustion of methane. -3169.o kJ

Example 8.8 The following equation represents the reaction between hydrochloric acid and a solution of sodium carbonate: 2H+ (aq) + CO32- (aq)  CO2 (g) + H2O (l) Calculate ΔH for this thermochemical equation. Ans: -2.2 kJ

Bond Enthalpy Bond enthalpy is defined as ΔH when one mole of bonds is broken in the gaseous state. H2 (g)  2H (g) ΔH = +436 kJ The above equation means that the ΔHf for 1 mole of hydrogen = 218 kJ Using bond enthalpies it is possible to explain why gas phase reactions are endothermic. Generally, a reaction is expected to be endothermic if: The bonds in the reactants are stronger than those in the products. There are more bonds in the reactants than in the products. -Generally enthalpies of formation are used rather than bond enthalpies because they are more accurate (within + or – 0.1 kJ).

1st Law of Thermodynamics Energy can be converted from one form to another but cannot be created nor destroyed. The change in energy is always energy transfer from system to surroundings, or vice versa. ΔEsystem = -ΔEsurroundings The total change in a system’s internal energy is the sum of the energy transferred as heat and/or work: ΔE = q + w

Example 8.9 Calculate ΔE of a gas for a process in which the gas: Absorbs 20 J of heat and does 12 J of work by expanding. Ans: +8 J Evolves 30 J of heat and has 52 J of work done on it as it contracts. Ans: +22 J

Enthalpy vs Energy ΔH = ΔE + ΔngRT R = 8.31 J/ mol K Δng = n gaseous products – n of gaseous reactants This equation is derived from the following, keeping in mind that work for a gas is is PV: H = E + PV * In this relationship H is really the q in the first law and PV is the w in the first law.

Example 8.10 Calculate ΔH and ΔE at 25 *C for the reaction that takes place when an oxyacetylene torch is used: C2H2 (g) + 5/2 O2 (g)  2CO2 (g) + H2O (g) Ans: ΔH = -1255.5 kJ and ΔE = -1254.3 kJ