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Standard entropies of X_{2}, Y_{2} and XY_{3} are 60, 40 and 50JK^{-1}mol^{-1} respectively. For the reaction

\frac{1}{2}X_{2}+\frac{3}{2}Y_{2}\leftrightharpoons , XY_{3},\Delta H=-30kJ

 

to be at equilibrium, the temperature should be:

  • Option 1)

    750K

  • Option 2)

    1000K

  • Option 3)

    1250K

  • Option 4)

    500K

 

Answers (1)

best_answer

 

Spontanous process -

\Delta G= \Delta H-\Delta T

 

- wherein

For spontanouse process \Delta G must be negative.

 

 given reaction is

\frac{1}{2}X^{2}+\frac{3}{2}Y^{2}\rightleftharpoons XY_{3}

We Know \Delta S^{0} =\sum S^{0} -\sum S^{0}rectant

 

= 50-\left ( 30+60 \right ) =-40 JK^{-1}mol^{-1}

at  equilibrium \Delta G^{0}=0

                       \Delta H^{0}=T\Delta S^{0}

\Rightarrow T= \frac{\Delta H^{0}}{\Delta S^{0}}=\frac{-30\times 10^{3}jmo^{-1}}{-40jk^{-1}mol^{-1}}

= 750 \:k


Option 1)

750K

Correct

Option 2)

1000K

Incorrect

Option 3)

1250K

Incorrect

Option 4)

500K

Incorrect

Posted by

Aadil

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