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?

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