Presentation on theme: "Thermochemistry Chapter 8 Energy is the capacity to do work Thermal energy is the energy associated with the random motion of atoms and molecules Chemical."— Presentation transcript:
Thermochemistry Chapter 8
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 Electrical energy is the energy associated with the flow of electrons Potential energy is the energy available by virtue of an objects position or composition.
Heat is the transfer of thermal energy between two bodies that are at different temperatures. Energy Changes in Chemical Reactions Temperature is a measure of the thermal energy. Temperature = Thermal Energy 90 0 C 40 0 C greater thermal energy
Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothing SYSTEM SURROUNDINGS
Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. energy, pressure, volume, temperature
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = ms Heat (q) absorbed or released: q = ms t q = C t t = t final - t initial
How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? s of Fe = J/g 0 C t = t final – t initial = 5 0 C – 94 0 C = C q = ms t = 869 g x J/g 0 C x –89 0 C= -34,000 J
Constant-Pressure Calorimetry No heat enters or leaves! q sys = q water + q cal + q rxn q sys = 0 q rxn = - (q water + q cal ) q water = ms t q cal = C cal t Reaction at Constant P H = q rxn
Constant-Volume Calorimetry No heat enters or leaves! q sys = q water + q bomb + q rxn q sys = 0 q rxn = - (q water + q bomb ) q water = ms t q bomb = C bomb t Reaction at Constant V H ~ q rxn H = q rxn
TWO Trends in Nature Order Disorder High energy Low energy
Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy energy + 2HgO (s) 2Hg (l) + O 2 (g) energy + H 2 O (s) H 2 O (l)
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. H = H (products) – H (reactants) H = heat given off or absorbed during a reaction at constant pressure H products < H reactants H < 0 H products > H reactants H > 0
Thermochemical Equations H 2 O (s) H 2 O (l) H = 6.01 kJ Is H negative or positive? System absorbs heat Endothermic H > kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm.
Thermochemical Equations CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) H = kJ Is H negative or positive? System gives off heat Exothermic H < kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm.
H 2 O (s) H 2 O (l) H = 6.01 kJ/mol ΔH = 6.01 kJ The stoichiometric coefficients always refer to the number of moles of a substance Thermochemical Equations If you reverse a reaction, the sign of H changes H 2 O (l) H 2 O (s) H = kJ If you multiply both sides of the equation by a factor n, then H must change by the same factor n. 2H 2 O (s) 2H 2 O (l) H = 2 mol x 6.01 kJ/mol = 12.0 kJ
H 2 O (s) H 2 O (l) H = 6.01 kJ The physical states of all reactants and products must be specified in thermochemical equations. Thermochemical Equations H 2 O (l) H 2 O (g) H = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s) H reaction = kJ 266 g P 4 1 mol P g P 4 x 3013 kJ 1 mol P 4 x = 6470 kJ
Standard enthalpy of formation ( H 0 ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero. H 0 (O 2 ) = 0 f H 0 (O 3 ) = 142 kJ/mol f H 0 (C, graphite) = 0 f H 0 (C, diamond) = 1.90 kJ/mol f
The standard enthalpy of reaction ( H 0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD H 0 rxn d H 0 (D) f c H 0 (C) f = [+] - b H 0 (B) f a H 0 (A) f [+] H 0 rxn H 0 (products) f = H 0 (reactants) f - Hesss Law: 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. (Enthalpy is a state function. It doesnt matter how you get there, only where you start and end.)
Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H 0 rxn H 0 (products) f = H 0 (reactants) f - H 0 rxn 6 H 0 (H 2 O) f 12 H 0 (CO 2 ) f = [+] - 2 H 0 (C 6 H 6 ) f  H 0 rxn = [ 12 × × ] – [ 2 × ] = kJ kJ 2 mol = kJ/mol C 6 H 6
Calculate the standard enthalpy of formation of CS 2 (l) given that: C (graphite) + O 2 (g) CO 2 (g) H 0 = kJ rxn S (rhombic) + O 2 (g) SO 2 (g) H 0 = kJ rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g) H 0 = kJ rxn 1. Write the enthalpy of formation reaction for CS 2 C (graphite) + 2S (rhombic) CS 2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C (graphite) + O 2 (g) CO 2 (g) H 0 = kJ 2S (rhombic) + 2O 2 (g) 2SO 2 (g) H 0 = x2 kJ rxn CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g) H 0 = kJ rxn + C (graphite) + 2S (rhombic) CS 2 (l) H 0 = (2x-296.1) = 86.3 kJ rxn 6.6
Chemistry in Action: Fuel Values of Foods and Other Substances C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) H = kJ/mol 1 cal = J 1 Cal = 1 kcal = 1000 cal = 4184 J
The enthalpy of solution ( H soln ) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. H soln = H soln - H components Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack?
The Solution Process for NaCl H soln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol
Energy Diagrams ExothermicEndothermic (a)Activation energy (Ea) for the forward reaction (b)Activation energy (Ea) for the reverse reaction (c) Delta H 50 kJ/mol300 kJ/mol 150 kJ/mol100 kJ/mol -100 kJ/mol+200 kJ/mol
Sublimation Deposition H 2 O (s) H 2 O (g) Molar heat of sublimation ( H sub ) is the energy required to sublime 1 mole of a solid. H sub = H fus + H vap ( Hesss Law)
Molar heat of fusion ( H fus ) is the energy required to melt 1 mole of a solid substance.
Sample Problem How much heat is required to change 36 g of H 2 O from -8 deg C to 120 deg C? Step 1: Heat the iceQ=mcΔT Q = 36 g x 2.06 J/g deg C x 8 deg C = J = 0.59 kJ Step 2: Convert the solid to liquidΔH fusion Q = 2.0 mol x 6.01 kJ/mol = 12 kJ Step 3: Heat the liquidQ=mcΔT Q = 36g x J/g deg C x 100 deg C = J = 15 kJ
Sample Problem How much heat is required to change 36 g of H 2 O from -8 deg C to 120 deg C? Step 4: Convert the liquid to gasΔH vaporization Q = 2.0 mol x kJ/mol = 88 kJ Step 5: Heat the gasQ=mcΔT Q = 36 g x 2.02 J/g deg C x 20 deg C = J = 1.5 kJ Now, add all the steps together 0.59 kJ + 12 kJ + 15 kJ + 88 kJ kJ = 118 kJ
Bond Enthalpies (Bond Energy) Strength of a chemical bond is measured by the bond enthalpy, H B Bond enthalpies are positive, because heat must be supplied to break a bond. Bond breaking is endothermic ( H is positive). Bond formation is exothermic ( H is negative). H 2 (g) --> 2 H H o = +436 kJ H B = 436 kJ/mol
Mean bond enthalpy: average molar enthalpy change accompanying the dissociation of a given type of bond.
Estimate the enthalpy change of the reaction between gaseous iodoethane and water vapor. CH 3 CH 2 I(g) + H 2 O(g) --> CH 3 CH 2 OH(g) + HI(g) Reactant: break a C-I bond and an O-H bond H o = 238 kJ kJ = 701 kJ Product: to form a C-O bond and an H-I bond H o = -360 kJ kJ = -659 kJ Overall enthalpy change = 701 kJ kJ = 42 kJ