The enthalpy of combustion (at 25 ?C), Delta H degree comb, for benzoic acid, C_

Question

The enthalpy of combustion (at 25 ?C), Delta H degree comb, for benzoic acid, C_7 H_6 O_2 (s) is - 3227 kJ mol^-1 and a) write the balanced chemical equation that describes the combustion of one mol of benzoic acid. b) Write the balanced chemical equation that describes the standard formation of benzoic acid. c) Using the information given above and the equations in parts a) and b), calculate Delta H_f degree (kJ mol^-1) for benzoic acid. d) Calculate Delta S_f degree (J mol^-1 k^-1) for benzoic acid and Delta S_rxn degree (J mol^-1 K^-1) for the combustion of one mole of benzoic acid. e) Calculate Delta G_f degree (kJ mol^-1) for benzoic acid and Delta G_rxn degree (kJ mol^-1) for the combustion of one mole of benzoic acid.

Explanation / Answer

SOLUTION:

(a) Balanced chemical equation for the combustion of one mole of benzoic acid

C7H6O2(s) + 15/2 O2(g) -----> 3H2O(l) + 7CO2(g) Hcomb = -3227KJ

(b)   3H2O(l) + 7CO2(g) -----> C7H6O2(s) + 15/2 O2(g)   Hr = 3227KJ

(c) Hr =   Hf (products) - Hf (Recatants)

Hf (products) = enthalpy of formation of products

Hf (rectants) = enthalpy of formation of reactants

3227KJ = {Hf C7H6O2(s) + Hf O2(g)} - {Hf H2O(l) + Hf CO2(g)}

3227KJ = {Hf C7H6O2(s) + 0} - {- 3 X 285.8KJ - 7 X 393.5KJ}

Hf C7H6O2(s) = 3227KJ - 3611.9 KJ = - 384.9KJ(d)

Note: The text of rest of the questions is illegible so cant be solved accurately.

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