Thermochemistry

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) 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

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The standard enthalpy of reaction (  H 0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD H0H0 rxn d  H 0 (D) f c  H 0 (C) f = [+] - b  H 0 (B) f a  H 0 (A) f [+] H0H0 rxn  H 0 (products) f =   H 0 (reactants) f  Hess’s 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 doesn’t 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) H0H0 rxn  H 0 (products) f =   H 0 (reactants) f  - H0H0 rxn 6  H 0 (H 2 O) f 12  H 0 (CO 2 ) f = [+] - 2  H 0 (C 6 H 6 ) f [] H0H0 rxn = [ 12 × × ] – [ 2 × ] = kJ kJ 2 mol = kJ/mol C 6 H 6 6.6

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

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

First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed.

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 6.5

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

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure. The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm Phase Changes

The critical temperature (T c ) is the temperature above which the gas cannot be made to liquefy, no matter how great the applied pressure. The critical pressure (P c ) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature.

Melting Freezing H 2 O (s) H 2 O (l) The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

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 ( Hess’s 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