JEE MAIN - Chemistry (2008 - No. 8)
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 :
$${1 \over 2} X_2$$ + $${3 \over 2} Y_2 \to$$ XY3, $$\Delta H$$ = -30 kJ, to be at equilibrium, the temperature will be :
1250 K
500 K
750 K
1000 K
Explanation
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 $$\,\,\,\,\,$$ = $$750$$ $$K$$
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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