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16 December, 21:31

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?

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  1. 16 December, 22:15
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    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
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