Standard entropies of X_2, Y_2 and XY_3 are 60,40 and 50 ~J ~K ~K^ -1 ~mol^ -1 respectively. For the reaction (1)/(2) X_2+(3)/(2) Y_2 leftharpoons XY_3 ; Δ H=-30 ~kJ , to be at equilibrium, the temperature should be

Correct answer: A. 750 ~K For the reaction, aligned & (1)/(2) X_ 2 +(3)/(2) Y _ 2 leftharpoons XY_ 3 ; Δ H=-30 kJ & Δ S^ ^ =S_ (XY_3) ^ ^ -[(1)/(2) S_ X_2 ^ ^ +(3)/(2) S_ Y_2 ^ ^ ] & =50-[(1)/(2) ×₆₀⁺(3)/(2) ×40] & =50-[30+60]=50-90 & =-40 JK^ -1 mol^ -1 aligned We know that, Δ G^ ^ =Δ H^ ^ -T Δ S^ ^ At equilibrium, aligned Δ G^ ^ & =0 Δ H& = T Δ S^ ^ T& = Δ H Δ S^ ^ = -30 × 10^3 J -40 J K^ -1 mol^ -1 =750 K aligned

NEET 2010

Exam: NEET
Year: 2010
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2010 · Chemistry · Thermodynamics (C)

Standard entropies of X_2, Y_2 and XY_3 are 60,40 and 50 ~J ~K ~K^...

Standard entropies of

\mathrm{X}_2, \mathrm{Y}_2

and

\mathrm{XY}_3

are 60,40 and

50 \mathrm{~J} \mathrm{~K} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

respectively. For the reaction

\frac{1}{2} \mathrm{X}_2+\frac{3}{2} \mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_3 ; \Delta \mathrm{H}=-30 \mathrm{~kJ}

, to be at equilibrium, the temperature should be

Options

A $750 \mathrm{~K}$
B $1000 \mathrm{~K}$
C $1250 \mathrm{~K}$
D $500 \mathrm{~K}$

Correct answer

A. $750 \mathrm{~K}$

Step-by-step solution

For the reaction,

\begin{aligned}& \frac{1}{2} \mathrm{X}_{2}+\frac{3}{2} \mathrm{Y}}_{2} \rightleftharpoons \mathrm{XY}_{3} ; \Delta \mathrm{H}=-30 \mathrm{ kJ}\\ & \Delta \mathrm{S}^{{}^{\circ}}=\mathrm{S}_{\left(\mathrm{XY}_3\right)}^{{}^{\circ}}-\left[\frac{1}{2} \mathrm{ S}_{\mathrm{X}_2}^{{}^{\circ}}+\frac{3}{2} \mathrm{ S}_{\mathrm{Y}_2}^{{}^{\circ}}\right]\\ & =50-\left[\frac{1}{2} \times₆₀⁺\frac{3}{2} \times40\right] \\ & =50-[30+60]=50-90 \\ & =-40 \mathrm{JK}^{-1} \mathrm{ mol}^{-1} \end{aligned}

We know that, \Delta \mathrm{G}^{{}^{\circ}}=\Delta \mathrm{H}^{{}^{\circ}}-\mathrm{T} \Delta \mathrm{S}^{{}^{\circ}}At equilibrium,

\begin{aligned} \Delta \mathrm{G}^{{}^{\circ}}& =0 \\ \Delta \mathrm{H}& = \mathrm{T} \Delta \mathrm{S}^{{}^{\circ}}\\ \mathrm{T}& =\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}^{{}^{\circ}}}=\frac{-30 \times 10^3 \mathrm{ J}}{-40 \mathrm{ J} \mathrm{ K}^{-1} \mathrm{ mol}^{-1}}=750 \mathrm{ K} \end{aligned}

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