QJEEM08Ch1b3e: JEE Main 2008 Chemistry - Class 11 JEEM08Ch1b3e Solution: Standard entropy of X 2 , Y 2 and XY 3 are 60, 40 and 50 JK −1…

Question

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Standard entropy of X₂, Y₂ and XY₃ are 60, 40 and 50 JK⁻¹ mol⁻¹ , respectively. For the reaction,

{1 \over 2} X_2

+

{3 \over 2} Y_2 \to

XY₃,

\Delta H

= -30 kJ, to be at equilibrium, the temperature will be :

Accepted Answer

For a reaction to be at equilibrium

\Delta G = 0.

Since

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

so at equilibrium

\Delta H - T\Delta S = 0

or

\,\,\,\,\Delta H = T\Delta S

For the reaction

{1 \over 2}{X_2} + {3 \over 2}{Y_2} \to X{Y_3};\,\,\,

\Delta H = - 30kJ\,\,\left( {given} \right)

Calculating

\Delta S

for the above reaction, we get

\Delta S = 50 - \left[ {{1 \over 2} \times 60 + {3 \over 2} \times 40} \right]J{K^{ - 1}}

= 50 - \left( {30 + 60} \right)J{K^{ - 1}}

= - 40\,J{K^{ - 1}}

At equilibrium,

\,\,\,T\Delta S = \Delta H

[ as

\,\,\,\Delta G = 0

]

\therefore

\,\,\,\,\,

T \times \left( { - 40} \right) = - 30 \times 1000

[ as

1kJ=1000J

]

or

\,\,\,\,\,

T = {{ - 30 \times 1000} \over { - 40}}

or

\,\,\,\,\,

=

750

K

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