Question

For hypothetical reversible reaction:

$\frac{1}{2}{A}_2\left(g\right)+\frac{3}{2}{B}_2\left(g\right)⟶A{B}_3\left(g\right);\Delta H=-20kJ$

The standard entropies of $A_2,B_2,$ and $A{B}_3$ are $60,40,$ and $50J{K}^{-1}mo{l}^{-1}$, respectively. The above reaction will be equilibrium at:

• A. 500 K
• B. 400 K
• C. 200 K
• D. 100 K

Answer 1

The correct option is A 500 K

As we know,
At equilibrium,$\Delta S_{total}=0$

$\therefore S_P^{\ominus }-S_R^{\ominus }=0$

$\Delta S_{\left(A{B}_3\right)}-\frac{1}{2}\Delta S_{\left(A_2\right)}-\frac{3}{2}\Delta S_{\left(B_2\right)}$

Also,

$\Delta G=\Delta H-T\Delta S=0$ at equilibrium,

$\Delta H=-20kJ=-20000J$

Therefore,

$\Delta H=T\Delta S$

$-20,000=T\times (50-30-60)$

$T=500K$

Hence, option A is correct.