Enthalpies of Reaction

Enthalpy change (H) for a chemical reaction is called

Hrxn and is given by

Hrxn = H(products) - H(reactants)

Notice: H(prod) and H(react) can’t be measured individually, but Hrxn is measured by measuring heat flow at constant P!!!

Notice:

Hrxn > 0: reaction endothermic; system absorbs heat

Hrxn < 0: reaction exothermic; system releases heat to surroundings

N2(g) + 3H2(g)  2NH3(g) Hrxn = -46.19 kJ/mol

104 Important features of H

H is an extensive property - its magnitude is directly proportional to amount of reactant consumed, e.g.,

2SO2(g) + O2(g)  2SO3(g) Hrxn = -196 kJ

What is Hrxn for the reaction of 4 mol of SO2(g) with 2 mol of O2(g)?

What is Hrxn for the reaction of 2.4 g of SO2(g)?

105 Hrxn for a reaction is equal in magnitude but opposite in sign to Hrxn for the reverse reaction, e.g.,

2SO3(g)  2SO2(g) + O2(g) Hrxn = 196 kJ H for a reaction depends on the state of the reactants and products (i.e., solid, liq., gas)

The states of the reacting species must be specified - we will assume that all are at 25o C

How do we determine H values experimentally?

Calorimetry Measurement of heat flow Heat capacity C: energy required to raise the temperature of an object by 1oC

C determines the temperature change a body experiences when it absorbs heat

The greater the heat capacity, the greater the heat required to produce a temperature increase

106 For pure substances, the heat capacity is usually given for a specified amount of substance

Heat capacity of 1 mol of a substance: molar heat capacity

Heat capacity of 1 g of a substance: specific heat

E.g., the specific heat of water is 4.184 J/g-K Calculate the molar heat capacity of water

Notice these units of the specific heat: if we multiply a specific heat x grams x T, we get units of heat

Therefore, heat flow q can be calculated from

q = (specific heat) x (mass) x T

107 E.g., How many kJ of heat are required to raise the temperature of 2.56 kg of water from 44.8oC to 92.0oC?

Calorimetric measurement of heats of reaction are usually done in one of two ways: constant pressure or constant volume Constant Pressure Calorimetry Works well for reactions in solution Easy to hold pressure constant – perform reaction in an insulated vessel open to atmosphere Assume that reaction vessel (calorimeter) neither absorbs nor loses heat – all heat produced (or absorbed) comes from the reaction being studied

How to calculate Hrxn from a constant-pressure calorimetric measurement?

108 E.g., When 4.25 g of NH4NO3 is dissolved in 60.0 g H2O in a 'coffee cup' calorimeter, the temperature drops from 22.0 oC to 16.9 oC. Find H for the solution process. Assume that the heat capacity of the solution is 4.184 J/g-K.

Hess's Law

H is defined in terms of state functions, so H is a state function H depends only on initial and final states......

Suppose we want to calculate Hrxn for the reaction A(g) + B(g)  C(g)

The initial state is A(g, P, 25oC) + B(g, P, 25oC) The final state is C(g, P, 25oC)

109 The enthalpy change Hrxn depends only on the enthalpy difference between the final state and the initial state, and not how we get from one to the other......

Suppose that we know Hrxn for the following reactions.....

A(g) + B(g)  E(g) Hrxn = 52 kJ/mol

C(g)  E(g) Hrxn = -105 kJ/mol

Can these two known reactions be combined to give the reaction that we want? (i.e., A(g) + B(g)  C(g)) (assume that all reactants and products are at pressure P and 25oC)

Question: are all reactants and products in the reaction we want contained in the reactions we know?

Question: are there any species in the known reactions that aren’t present in the reaction we want?

How to make these species go away?

What happens to the sign of Hrxn if we reverse a reaction?

110 Hess’s Law tell us that if a reaction can be carried out in a series of steps, then....

The sum of the H for the individual steps must = H for the overall process

In the above example, instead of going directly from A&B to C, we convert A&B to E, and then E to C

As long as we start with A&B and end up with C, it doesn’t matter how we get there!!!

E.g., from the following heats of reaction,

N2(g) + 2O2(g) 2NO2(g) Hrxn = 67.6 kJ

2NO(g) + O2(g) 2NO2(g) Hrxn = -56.6 kJ

Find the heat of reaction Hrxn for

N2(g) + O2(g)2NO(g)

111 Can we tabulate Hrxn values that would allow us to use Hess's Law to calculate enthalpy changes for common reactions?

We can’t tabulate Hrxn values for every single reaction known to science – some reactions are too dangerous or too difficult to attempt.....

Also: Hrxn depends on temperature and pressure – tabulating heats of reaction for every single reaction at different values of T and P would be rather nightmarish

How to get around these difficulties?

Heats of Formation

Define Hf : heat of formation

Hf is the enthalpy change associated with the formation of a compound from its constituent elements

REMEMBER: H depends on states of products and reactants

We define Ho as a standard enthalpy change - takes place with all elements in their standard states Standard state: most stable form of a substance at 1 atm pressure

112 So.... we can have

o H vap = standard heat of vaporization

o H fus = standard heat of fusion The “o” means ‘standard’, which means 1 atm pressure Standard enthalpies are generally tabulated at 25oC (298 K)

o E.g., What does H rxn mean?

o So... H f is a standard heat of formation Enthalpy change for the formation of 1 mol of substance from its elements, all substances in their standard states

KNOW the common standard states (e.g., O2(g),

C(graphite), H2(g))

o By definition, H f for the most stable form of an element = 0 kJ (think about this – write the formation reaction for C(graphite))

113 o H f values are tabulated in Appendix C, e.g.,

AlCl3(s) -705.6 kJ/mol

CO2(g) -393.5 kJ/mol

NH3(g) -46.19 kJ/mol

E.g. write the formation reactions for AlCl3(s), CO2(g), and NH3(g)

o Now: can we use Hess's Law and tabulated H f values to calculate enthalpy changes for reactions?

o H rxn : standard enthalpy change for a reaction is given by

o o o H rxn =  nH f(products) -  mH f(reactants)

Where n and m are the stoichiometric coefficients from the reaction......

114 o E.g., given the following H f ,

NH4NO3(s) -356.6 kJ/mol

N2O(g) 81.6 " "

H2O(g) -241.8 " "

o Calculate H rxn for

NH4NO3(s)  N2O(g) + 2H2O(g)

115 Problems du Jour

From the following enthalpies of reaction:

H2(g) + F2(g)  2HF(g) -537 kJ

C(s) + 2F2(g)  CF4(g) -680 kJ

2C(s) + 2H2(g)  C2H4(g) 52.3 kJ

Find Hrxn for the reaction

C2H4(g) + 6F2(g)  2CF4(g) + 4HF(g)

116 ProblemduJour

Complete combustion of 1 mol of acetone, C3H6O, results in the liberation of 1790 kJ of heat:

o C3H6O(g) + 4O2(g)  3CO2(g) + 3H2O(l) H rxn = -1790 kJ

Using this information, along with the following thermochemical data:

o Substance H f, kJ

CO2(g) -393.5

H2O(l) -285.83

o Calculate H f for acetone.

117 Problems du Jour

o Using data from Appendix C, calculate H rxn for the following reactions:

2KOH(s) + CO2(g)  K2CO3(s) + H2O(g)

Fe2O3(s) + 6HCl(g)  2FeCl3(s) + 3H2O(g)

118 Problem du Jour

When a 9.55 g sample of solid sodium hydroxide dissolves in 100.0 g of water in a coffee-cup calorimeter, the temperature rises from 23.6 oC to 47.4 oC. Calculate H (in kJ/mol NaOH) for the solution process NaOH(s)  Na+(aq) + Cl-(aq) Assume that the specific heat of the solution is the same as that of pure water.