Question

For a reversible reaction :
$X_{(g)}+3Y_{(g)}\leftrightharpoons 2Z_{(g)}$
$\Delta H=-40kJ$ the standard entropies of $X,Y$ and $Z$ are $60,40$ and $50J{K}^{-1}mo{l}^{-1}$ respectively.
The temperature at which the above reaction attains equilibrium is about :

• A. 400 K
• B. 500 K
• C. 273 K
• D. 373 K

Answer 1

Answer: (B)

$X(g)+3Y(g)2Z(g)$

$\Delta S^{\circ }=2{\bar{S}}^{\circ }(Z)-\left\{S^{\circ }(X)+3S^{\circ }(Y)\right\}$

$=2\times 50-\{60+3\times 40\}$

$=100-180=-80J{K}^{-1}mo{l}^{-1}$

Given $\Delta H^{\circ }=-40kJ=-40,000J$

$\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$

At equilibrium, $\Delta G^{\circ }=0$

$\therefore \Delta H^{\circ }=T\Delta S^{\circ }$

or $T=\frac{\Delta H^{\circ }}{\Delta S^{\circ }}=\frac{40,000}{80}=500K$