Standard entropies of X2, Y2 and XY5 are 70, 50 and 110 J K^{–1} mol^{–1} respectively.

Question

Standard entropies of \(X_{2},\ Y_{2}\) and \(X Y_{5}\) are 70, 50 and \(110 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) respectively. The temperature in Kelvin at which the reaction

\(\frac{1}{2} \mathrm{X}_{2}+\frac{5}{2} \mathrm{Y}_{2} \rightarrow \mathrm{XY}_{5} \ \Delta \mathrm{H}^{-}=-35 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Will be at equilibrium is _____ (Nearest integer)

Answer 1

Answer is : 700

\(\frac{1}{2} X_{2}+\frac{5}{2} Y_{2} \rightleftharpoons X Y_{5}\) 

\( \Delta \mathrm{S}_{\mathrm{Rxn}}^{0}=110-\left[\left(\frac{1}{2} \times 70\right)+\left(\frac{5}{2} \times 50\right)\right]\) 

\( =110-160=-50 \ \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) 

\(\Delta \mathrm{G}^{0}=0\) at eqb

\( \Delta \mathrm{G}^{0}=\Delta \mathrm{H}^{0}-\mathrm{T} \Delta \mathrm{S}^{0}\) 

\( 0=-35000-\mathrm{T}(-50)\) 

\( \mathrm{T}=700\ \text{Kelvin}\)