Question

The ${\Delta }_f{H}^o$ for the $C{O}_2\left(g\right),CO\left(g\right)$ and $H_{2}O\left(l\right)$ are $-393.5,-110.5$ and $-241.8{\text{kJ mol}}^{-1}$ respectively. The standard internal energy change (in ${\text{kJ mol}}^{-1}$) for the given reaction at $298\text{K}$ is:
$C{O}_2\left(g\right)+H_2\left(g\right)\to CO\left(g\right)+H_{2}O\left(l\right)$
(Given $R=8.3\text{J/K/mol}$)

• A. $42.1$
• B. $43.67$
• C. $-43.67$
• D. $-42.1$

Answer 1

The correct option is B $43.67$

First write the balanced chemical equation for the formation of carbon dioxide, carbon monoxide and water :

$C\left(s\right)+O_2\left(g\right)\to C{O}_2\left(g\right);\Delta H^o=-393.5{\text{kJ mol}}^{-1}....\left(i\right)$
$C\left(s\right)+\frac{1}{2}{O}_2\left(g\right)\to CO\left(g\right);\Delta H^o=-110.5{\text{kJ mol}}^{-1}....\left(ii\right)$
$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(l\right);\Delta H^o=-241.8{\text{kJ mol}}^{-1}....\left(iii\right)$

then obtain the required equation by (ii) + (iii) - (i)
$C{O}_2\left(g\right)+H_2\left(g\right)\to CO\left(g\right)+H_{2}O\left(l\right)\Delta H^o=+41.2{\text{kJ mol}}^{-1}$

Hence, the standard enthalpy of the reaction is $+41.2{\text{kJ mol}}^{-1}$.
$\Delta H=\Delta U+\Delta nRT$
$\Delta n=1-2=-1$
$\Delta U=\Delta H+RT$
$\Delta U=41200+2473.4=43673.4\text{J}$
$\Rightarrow 43.67{\text{kJ mol}}^{-1}$