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Given that $\delta H_{r298K}= -54.07kJ \:mol^{-1}$ and $\delta S^{\circ}_{r298K}= 10J\:mol^{-1}$ and $R= 8.314JK^{-1}\:mol^{-1}$ . The value of log₁₀K for a reaction, $A\leftrightarrow B$ is:

Option 1)
5

Option 2)
10

Option 3)
95

Option 4)
100

Answers (1)

As learnt in

? G of equilibrium -

$\Delta G_{0}= -2.303 RT \log K_{c}$

- wherein

At Equilibrum

$\Delta G= 0$

and $Q= K_{c}$

and

Gibb's free energy (? 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

$\Delta G^{\circ}= \Delta H^{\circ} - T\Delta S^{\circ}$

$=-54.07 \times 10^{3}-298\times 10$

$=-57050$

$\Delta G^{\circ}= -2.303\ RT \log k$

$\log k = \frac{57050}{2.303\times 8.314\times 298} = \frac{57050}{5705}= 10$

Option 1)

5

This option is incorrect

Option 2)

10

This option is correct

Option 3)

95

This option is incorrect

Option 4)

100

This option is incorrect

Posted by

Vakul

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