If $K_{1}$ and $K_{2}$ are the equilibrium constants at temperature $T_{1}$ and $T_{2}$ , where $T_{2} > T_{1}$ , then :

K_{2} = K

when

Δ H

is 0

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K_{2} > K_{1}

when

Δ H

is +ve

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K_{2} < K_{1}

when

Δ H

is -ve

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K_{2} < K_{1}

when

Δ H

is +ve

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Solution

The correct options are
A $K_{2} = K$ when $Δ H$ is 0
B $K_{2} > K_{1}$ when $Δ H$ is +ve
C $K_{2} < K_{1}$ when $Δ H$ is -ve
The van't Hoff equation is $l o g \frac{K_{2}}{K_{1}} = + \frac{Δ H}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}] => \frac{Δ H}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$ .
Here, $K_{1}$ and $K_{2}$ are the values of the equilibrium constants at temperatures $T_{1}$ and $T_{2}$ respectively.
$Δ H$ is the enthalpy of the reaction and R is the ideal gas constant.
When $Δ H = 0$ and $T_{2} > T_{1}$ , then term $\frac{Δ H}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$ is zero.
Hence, $K_{2} K_{1} = 1$ or $K_{2} = K_{1}$
When $Δ H > 0$ and $T_{2} > T_{1}$ , then the term $\frac{Δ H}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}] > 0$ .
Hence, $K_{2} K_{1} > 1$ or $K_{2} > K_{1}$
When $Δ H < 0$ and $T_{2} > T_{1}$ , then the term $\frac{Δ H}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}] < 0$ .
Hence, $K_{2} K_{1} < 1$ or $K_{2} < K_{1}$

Suggest Corrections

Similar questions

Q. Assertion :Th equilibrium constant of the exothermic reaction at high temperature decreases. Reason: Since

ln \frac{K_{2}}{K_{1}} = \frac{{Δ H}^{o}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]

and for exothermic reaction,

{Δ H}^{o} = - v e

and thereby;

\frac{K_{2}}{K_{1}} < 1

Two blocks of the same thickness and thermal conductivities $k_{1} a n d k_{2}$ are placed in contact. Their outer surfaces are kept at constant temperatures $T_{1} a n d T_{2}$ . If $T_{2}$ > $T_{1}$ , and $k_{2}$ > $k_{1}$ the temperature of the interfaces in the steady state will be