Question

If $S^0$ for $H_2$, ${Cl}_2$ and HCl are 0.13, 0.22 and 0.19 kJ $K^{-1}{mol}^{-1}$ respectively. The total change in standard entropy for the reaction $H_2+{Cl}_2⟶2HCl$ is:

• A. 30 ${JK}^{-1}{mol}^{-1}$
• B. 40 ${JK}^{-1}{mol}^{-1}$
• C. 60 ${JK}^{-1}{mol}^{-1}$
• D. 20 ${JK}^{-1}{mol}^{-1}$

Answer 1

The correct option is A 30 ${JK}^{-1}{mol}^{-1}$

$H_2+{Cl}_2⟶2HCl$

${\Delta S^0}_{H_2}=0.13KJ/mol-K$

${\Delta S^0}_{{Cl}_2}=0.22KJ/mol-K$

${\Delta S^0}_{HCl}=0.19KJ/mol-K$

$\Delta S={\Delta S^0}_{product}-{\Delta S^0}_{reactant}$

$\Delta S=(2\times 0.19)-(0.13+0.22)$

$\Rightarrow \Delta S=0.38-0.35=0.03KJ/mol-K=30J/mol-K$

Hence the total change in standard entropy for the given is $30J/mol-K$.