VITEEE 2016 Paper

For a reaction to be at equilibrium ΔG = 0. Since ΔG = ΔH - TΔS so at equilibrium ΔH -TΔS = 0 or ΔH = TΔS For the reaction $\frac{1}{2} X_2 + \frac{3}{2} Y_2$ → $XYY_3$ ; ΔH = - 30 kJ Calculating AS for the above reaction, we get ΔS = 50 - $\left[ \frac{1}{2} \times 60 + \frac{3}{2} \times 40 \right]$ $\text{JK}^{-1}$ = 50 - (30 + 60) $\text{JK}^{-1}$ = - 40 $\text{JK}^{-1}$ At equilibrium, TΔS = ΔH [Since ΔG = 0] ∴ T × (- 40) = - 30 = 1000 [Since 1 kJ = 1000 J] or T = $\frac{-30 \times 1000}{-40}$ or 750 K