For hypothetical reversible reaction

$\frac{1}{2}{\mathrm{A}}_2\left(\mathrm{g}\right)+\frac{3}{2}{\mathrm{B}}_2\left(\mathrm{g}\right) \to {\mathrm{AB}}_3\left(\mathrm{g}\right)$; $\mathrm{\Delta H}$ is – 20kJ. Standard entropies of ${\mathrm{A}}_2$ , ${\mathrm{B}}_2$  and ${\mathrm{AB}}_3$  are 60, 40 and 50J/k mol respectively. Then temperature of reaction at equilibrium is

Given data: $\Delta H$ is – 20kJ.

$A_2$ , $B_2$  and $A{B}_3$  are 60, 40 and 50J/k mol respectively.

Formula/Concept used: $\Delta G=\Delta H-T\Delta S$ 

Detailed solution: The reaction given is:

$\frac{1}{2}{A}_2\left(g\right)+\frac{3}{2}{B}_2\left(g\right) \to A{B}_3\left(g\right)$

To calculate the change in entropy will be:

$\mathrm{\Delta S}={\sum \mathrm{\Delta S}}_{\mathrm{prod}}-{\sum \mathrm{\Delta S}}_{\mathrm{reac}}$ 

$$\begin{gathered} \Delta S=50-(\frac{1}{2}\times 60+\frac{3}{2}\times 40) \\ \Delta S=-40\text{ J/K mol} \end{gathered}$$

To calculate the temperature, we can use:

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

At equilibrium, $\Delta G=0$.

Putting the values, we get:

$$\begin{gathered} \mathrm{\Delta H}=\mathrm{T\Delta S} \\ \mathrm{T}=\frac{\mathrm{\Delta H}}{\mathrm{\Delta S}}=\frac{-20\times 1000}{-40}=500\text{ K} \end{gathered}$$

Therefore, the correct option is B.