Get Answers to all your Questions

header-bg qa

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)


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}}


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

- wherein


\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


View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE