Thermodynamics And Thermochemistry Result Question 5
5. For the complete combustion of ethanol, $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)+3 \mathrm{O}_2(g) \longrightarrow 2 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l)$, the amount of heat produced as measured in bomb calorimeter, is $1364.47 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $25^{\circ} \mathrm{C}$. Assuming ideality the enthalpy of combustion, $\Delta_C H$, for the reaction will be ( $\left.R=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$
(2014 Main)
(a) $-1366.95 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(b) $-1361.95 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(c) $-1460.50 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(d) $-1350.50 \mathrm{~kJ} \mathrm{~mol}^{-1}$
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Answer:
Correct Answer: 5. (a)
Solution:
$\begin{aligned} & \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)+3 \mathrm{O}_2(g) \longrightarrow 2 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l) \\ & \Delta U=-1364.47 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta H=\Delta U+\Delta n_g R T \\ & \Delta n_g=-1 \\ & \Delta H=-1364.47+\frac{-1 \times 8.314 \times 298}{1000} \end{aligned}$
[Here, value of $R$ in unit of J must be converted into kJ ]
$\begin{aligned} & =-1364.47-2.4776=-1366.9476 \mathrm{~kJ} / \mathrm{mol} \\ \text { or } \quad & =-1366.95 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$
BETA
