One mole of \mathrm{C}_{8} \mathrm{H}_{14} and 14.0 mol of oxygen are placed into a lab scale batch reactor. The reactor then is placed into a water bath at 25^{\circ} \mathrm{C}; the
water bath contains 20.0 kg of water. The contents of the reactor are burned. The \mathrm{C}_{8} \mathrm{H}_{14} combusts completely to give \mathrm{H}_{2} \mathrm{O} and \mathrm{CO}_{2}. At the end of the combustion, the water bath and the reactor and its contents are at 90^{\circ} \mathrm{C}. Under these conditions, the water produced during the combustion can be assumed to be completely in its liquid form. The top surface of the batch reactor is insulated such that all heat lost from the reactor goes to the water bath. What is the standard state heat of combustion of the \mathrm{C}_{8} \mathrm{H}_{14} ?
The following reaction took place:
\mathrm{C}_{8} \mathrm{H}_{14}+\frac{23}{2} \mathrm{O}_{2} \rightarrow 8 \mathrm{CO}_{2}+7 \mathrm{H}_{2} \mathrm{O}

Passo 1

Resumindo o enunciado: temos um mol de \mathrm{C}_{8}\mathrm{H}_{14} queima completamente num reator, liberando calor que vai para um banho de 20{,}0 kg de água, que se aquece de 25{\,}^\circ\mathrm{C} até 90{\,}^\circ\mathrm{C}. Queremos a entalpia padrãode combustão de \mathrm{C}_{8}\mathrm{H}_{14} em \mathrm{kJ/mol}.

Passo 2

Aplicamos o balanço de energia para calorimetria: todo o calor liberado pela reação, \;Q_{\mathrm{rxn}}, é absorvido pela água do banho. Assim,
Q_{\mathrm{rxn}} + Q_{\mathrm{água}} = 0á
onde
Q_{\mathrm{água}} = m \, c \,\Delta Tá
com m=20{,}0 kg, c=4{,}184, \mathrm{kJ/(kg\cdot K)} e \Delta T = 90 - 25 =65,K.

Passo 3

Calculando o calor absorvido pela água:
Q_{\mathrm{água}} = 20{,}0\;\mathrm{kg}\;\times 4{,}184\;\mathrm{\tfrac{kJ}{kg\cdot K}}\;\times 65\;\mathrm{K} = 20 \times 4{,}184 \times 65 \approx 5\,439{,}2\;\mathrm{kJ}.á

Passo 4

Como a reação libera calor, temos
Q_{\mathrm{rxn}} = -\,Q_{\mathrm{água}} \approx -5\,439{,}2\;\mathrm{kJ}á
por mol de \mathrm{C}_{8}\mathrm{H}_{14}.

Resposta

\Delta H^\circ_{\mathrm{comb}}\bigl(\mathrm{C}_{8}\mathrm{H}_{14}\bigr) \approx -5\,439\;\mathrm{kJ/mol}.

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