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AIEEE 2008 Solved Paper

Section: Chemistry

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Question : 6 of 104

Marks: +1, -0

Standard entropy of

X_2, Y_2

X Y_3

50\,\mathrm{JK}^{-1}\,\mathrm{mol}^{-1}

, respectively. For the reaction,

\; \frac{1}{2} X_2 + \; \frac{3}{2} Y_2 \rightarrow \mathrm{XY}_3, \Delta H = -30 \;\mathrm{kJ}

, to be at equilibrium, the temperature will be

1250 K
500 K
750 K
1000 K

Solution:

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

\; \frac{1}{2} X_2 + \; \frac{3}{2} Y_2 \rightarrow X Y_3

\Delta H = -30 \;\mathrm{kJ}

(given

)

Calculating

\Delta S

for the above reaction, we get

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

=50-(30+60)\,\mathrm{J\,K}^{-1}

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

At equilibrium,

T \Delta S = \Delta H

[ as

\Delta G = 0

]

\therefore \;\; T \times (-40) = -30 \times 1000

[ as

1 \,\mathrm{kJ} = 1000 \,\mathrm{J}

]

or

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

or

\;\; 750 \,\mathrm{K}

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