Question

For reversible reaction: $X_{\left(g\right)}+3Y_{\left(g\right)}\rightleftharpoons 2Z_{\left(g\right)}$; $\Delta H=-40KJ$
Standard entropies of X, Y and Z are 60, 40 and 50 J $K^{-1}mo{l}^{-1}$ respecctively. The temperature at which the above reaction is in equilibrium is :

• A. 273 K
• B. 600 K
• C. 500 K
• D. 400 K

Answer 1

The correct option is C 500 K

For the reaction,

$X+3Y\rightleftharpoons 2Z$

$\Delta S=2\times 50-(60+3\times 40)=-80kJ$

Also, $\Delta G=\Delta H-T\Delta S$,

At Equilibrium, $\Delta G=0$

Hence, $T=\frac{\Delta H}{\Delta S}=\frac{-40\times 1000}{-80}=500K$

Hence, the correct answer is option $\text{C}$.