Question

Heat of formation of $H_{2}O$ (g) at ${25}^{0}C$ is -243 KJ. $\Delta U$ for the reaction $H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(g\right)$ at ${25}^{0}C$ is:

• A. 241.8 KJ
• B. -241.8 KJ
• C. -243 KJ
• D. 243 KJ

Answer 1

Answer: (B)

-241.8 KJ

Solution:- (B) $-241.8kJ$

${H_2}_{\left(g\right)}+\frac{1}{2}{O_2}_{\left(g\right)}⟶{H_{2}O}_{\left(g\right)}\Delta H=-243kJ$

For the above reaction-

$\Delta n_g=n_P-n_R=1-(1+\frac{1}{2})=\frac{-1}{2}$

Given:- $T=25℃=(25+273)=298K$

As we know that,

$\Delta H=\Delta U+\Delta n_{g}RT$

$\Rightarrow -243=\Delta U+(\frac{-1}{2}\times 8.314\times {10}^{-3}\times 298)$

$\Rightarrow \Delta U=-243+1.2=-241.8kJ$

Hence the value of $\Delta U$ for the given reaction is $-241.8kJ$.