Question

If $\Delta H$ for the following reaction is - 120 cal at ${27}^{o}C$ then the value of $\Delta U$ for the reaction is:
${SO}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to {SO}_3$

• A. 420 cal
• B. 180 cal
• C. 300 cal
• D. 280 cal

Answer 1

The correct option is B 180 cal

Given:-

$\Delta H=-120cal$

$T=27℃=(27+273)K=300K$

For the given reaction,

${S{O}_2}_{\left(g\right)}+\frac{1}{2}{O_2}_{\left(g\right)}⟶{S{O}_3}_{\left(g\right)}$

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

Now as we know that,

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

$\Rightarrow \Delta U=\Delta H-\Delta n_{g}RT$

$\Rightarrow \Delta U=-120-(-\frac{1}{2}\times 2\times 300)$

$\Rightarrow \Delta U=-120+300=180cal$

Hence the value of $\Delta U$ for the given reaction is $180cal$.