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Standard entropy of $X_{2},Y_{2}\; and\; XY_{3}$ are 60, 40 and 50 J K^-1 mol^-1, respectively. For the reaction,

$1/2\; X_{2}+3/2\; Y_{2}\rightarrow XY_{3},\; \Delta H=-30\, kJ,$ to be at equilibrium, the temperature will be

Option 1)
1000 K

Option 2)
1250 K

Option 3)
500 K

Option 4)
750 K

Answers (1)

best_answer

As we learnt in

Gibb's free energy (Δ G) -

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

- wherein

$\Delta G=$ Gibb's free energy

$\Delta H=$ enthalpy of reaction

$\Delta S=$ entropy

$T=$ temperature

Entropy for isothermal process -

$\Delta S= nR\ln \frac{V_{f}}{V_{i}}$

or

$\Delta S= nR\ln \frac{P_{i}}{P_{f}}$

- wherein

$T_{f}=T_{i}$

$\Delta T=0$

$\Delta S=nR\ ln\frac{V_2}{V_1}=2.303nR \:log \frac{V_2}{V_1}\\ =2.303\times 2\times 8.314\times log(\frac{100}{10})=38.3Jmol^{-1}K^{-1}$

For a reaction to be equilibrium $\Delta G=0$ since $\Delta G=\Delta H-T\Delta S$ so at equilibrium $\Delta H-T.\Delta S= 0$

$\Delta H=T.\Delta S$

For reaction

$\frac{1}{2}X_{2}+\frac{3}{2}Y_{2}\rightarrow XY_{3},\:\Delta H= 30kJ \:\:(given)$

Calculating $\Delta S$ for the above equation, we get

$\Delta S= 50-\left [ \frac{1}{2} \times 60+\frac{3}{2}\times 40\right ]Jk^{-1}$

At equilibrium, $T.\Delta S=\Delta H\:\:\:\:\left ( \because \Delta G=0 \right )$

$\therefore T\times \left ( -40 \right )= -30\times 1000\:\:\:\left ( \because 1kJ=1000J \right )$

$T=\frac{-30\times 1000}{-40}=750K$

Option 1)

1000 K

This option is incorrect

Option 2)

1250 K

This option is incorrect

Option 3)

500 K

This option is incorrect

Option 4)

750 K

This option is correct

Posted by

prateek

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