12. Suppose 0.562 g of graphite is placed in a calorimeter with an excess of oxygen at 25.00˚C and 1 atm pressure. On reaction, the calorimeter temperature rises from 25.00˚C to 25.89˚C. The heat capacity of the calorimeter and its contents was determined in a separate experiment to be 20.7 kJ/˚C. What is the heat of reaction at 25.00˚C and 1 atm pressure?

the heat of reaction is - 18.423 kJ (exothermic) or - 393.37 kJ/mol of graphite

Explanation:

If we neglect the absorption of heat of the excess of oxygen because is low compared with the calorimeter, then we can assume that all the heat released by the reaction Q gra is absorbed by the calorimeter Q cal (if we assume also that the calorimeter is completely isolated). Then

Q gra + Q cal = Q surr = 0

since

- Q gra = ΔU (constant volume process)

- Q cal = C * (T final - T initial), where C = heat capacity of the calorimeter, T final = final temperature of the calorimeter, T initial = initial temperature o the calorimeter,

ΔU + C * (T final - T initial) = 0 → ΔU = C * (T initial - T final)

but the relationship between constant volume heat (ΔU) and constant pressure heat (ΔH = heat of reaction) is

ΔH = ΔU + R*T*Δn

where Δn = variation of moles of gas due to the reaction.

Since the oxygen is in excess we assume that all the graphite will react and turn into carbon dioxide

C + O₂ → CO₂, but since the stoichiometric coefficients are the same Δn=0 (one mole of O₂ consumed generates one mole of CO₂, thus there is no variation in the number of moles of gas)

therefore

ΔH = ΔU = C * (T initial - T final) = 20.7 kJ/˚C * (25.00˚C - 25.89˚C) = - 18.423 kJ

or the specific heat of reaction

Δh = ΔH/m = 18.423 kJ / 0.562 gr * 12gr/mol = 393.37 kJ/mol of graphite