Question

The standard enthalpies of formation at $300\text{K}$ for $CC{l}_4\left(l\right),H_{2}O\left(g\right),C{O}_2\left(g\right)$ and $HCl\left(g\right)$ are $-107,-242,-394$ and $-93{\text{kJ mol}}^{-1}$ respectively. The value of $\Delta U$ for the following reaction at $300\text{K}$ is:
$CC{l}_4\left(l\right)+2H_{2}O\left(g\right)\to C{O}_2\left(g\right)+4HCl\left(g\right)$

• A. $-170{\text{kJ mol}}^{-1}$
• B. $-175{\text{kJ mol}}^{-1}$
• C. $-182.50{\text{kJ mol}}^{-1}$
• D. $-282.50{\text{kJ mol}}^{-1}$

Answer 1

The correct option is C $-182.50{\text{kJ mol}}^{-1}$

The reaction given is,
$CC{l}_4\left(l\right)+2H_{2}O\left(g\right)\to C{O}_2\left(g\right)+4HCl\left(g\right)$
So, $\Delta H=\Delta H_f(C{O}_2,g)+4\Delta H_f(HCl,g)-\Delta H_f(CC{l}_4,l)-2\Delta H_f(H_{2}O,g)$
$\Rightarrow \Delta H=(-394-4\times 93+107+2\times 242){\text{kJ mol}}^{-1}$
$\Rightarrow \Delta H=-175{\text{kJ mol}}^{-1}$
We know,
$\Delta H=\Delta U+\Delta n_{g}RT$
$\Rightarrow \Delta U=\Delta H-\Delta n_{g}RT$
$=-175-\left(3\right)(8.314\times {10}^{-3}{\text{kJ K}}^{-1}mol-1)\left(300\text{K}\right)\approx -182.48{\text{kJ mol}}^{-1}$