Question

The value of enthalpy change ($\Delta$H) for the reaction $C_{2}H_{5}OH_{\left(I\right)} + 3O_{2\left(g\right)} \to 2CO_{2\left(g\right)} + 3H_{2}O_{\left(I\right)}$ at $27^{\circ}C$ is $-1366.5 \,kJ \,mol^{-1}.$ The value of internal energy change for the above reaction at this temperature will be :

• A. $-1369.0\, kJ$
• B. $-1364.0 \,kJ$
• C. $-1361.5\, kJ$
• D. $-1371.5 \,kJ$

Answer 1

Answer: (B)

$-1364.0 \,kJ$

Answer 2

$C_{2}H_{5}OH\left(\ell\right) + 3O_{2\left(g\right)} \to 2CO_{2\left(g\right)} + 3H_{2}O\left(\ell\right)$ $\Delta n_{g} = 2-3 = -1$ $\Delta U = \Delta H - \Delta n_{g} RT$ $= -1366.5-\left(-1\right)\times\frac{8.314}{10^{3}}\times300$ $= - 1366.5 + 0.8314 ? 3 = - 1364 KJ$