Standard entropies of , and are 70, 50 and 110 respectively. The temperature in Kelvin at which the reaction will be at equilibrium is _____ (Nearest integer). [2025]

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Solution

Q.
Standard entropies of $X_{2}$ , $Y_{2}$ and $X Y_{5}$ are 70, 50 and 110 ${JK}^{- 1}$ ${mol}^{- 1}$ respectively. The temperature in Kelvin at which the reaction

$\frac{1}{2} X_{2} + \frac{5}{2} Y_{2} ⇌ X Y_{5}, Δ H^{⊝} = - 35 {kJ mol}^{- 1}$

will be at equilibrium is _____ (Nearest integer). [2025]

Ans.
(700)

$\frac{1}{2} X_{2} + \frac{5}{2} Y_{2} ⇌ X Y_{5}$

$Δ S ° = S_{products} - S_{reactants} = S_{X Y_{5}} - [\frac{1}{2} S_{X_{2}} + \frac{5}{2} S_{Y_{2}}]$

$= 110 {JK}^{- 1} {mol}^{- 1} - [\frac{1}{2} \times 70 + \frac{5}{2} \times 50] {JK}^{- 1} {mol}^{- 1}$

$= - 50 {JK}^{- 1} {mol}^{- 1}$

$Δ G ° = Δ H ° - T Δ S ° = - 35 {kJ mol}^{- 1} - T \times (- 50 {JK}^{- 1} {mol}^{- 1})$

$= - 35000 {J mol}^{- 1} + T \times 50 {JK}^{- 1} {mol}^{- 1}$

Taking $Δ G ° = 0$ for equilibrium, $- 35000 {J mol}^{- 1} + T \times 50 {JK}^{- 1} {mol}^{- 1} = 0$

$T = \frac{35000}{50} K = 700 K$

(Note: At equilibrium $Δ G = 0$ , $Δ G °$ is not necessarily zero at equilibrium, but as per data provided in the question, the only way to solve the question is by taking $Δ G ° = 0$ ).

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