Question

For the conversion of 1 mole of $S{O}_2\left(g\right)$ into $S{O}_3\left(g\right)$ the enthalpy of reaction at constant volume, $\Delta U$ at $298K$ is $-97.027KJ$. What is the enthalpy of reaction, $\Delta H$ at constant pressure?

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

From the explanation given I the question we can write the chemical reaction as.

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

$\Delta ng$ = number of moles of gaseous products - number of moles of gaseous reactants

$\Delta ng$ = $1-1-\frac{1}{2}=-\frac{1}{2}$

The change in enthalpy can be easily given as

$\Delta H=\Delta U+\Delta nRT$

$\Delta H=(-97.027)\times {10}^3+(-\frac{1}{2}\times 8.314\times 298)$

= $(-97027)+(-1238.786)$
$\Delta H=-98.267kJ$