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Vant Hoff's Equation and Effect of Temperature on Equilibrium Constant

The effect of...

Question

The effect of temperature on equilibrium constant is expressed as $(T_{2} > T_{1})$ $l o g \frac{K_{2}}{K_{1}} = \frac{- Δ H}{2.303} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$ . For endothermic reaction false statement is:

A

[\frac{1}{T_{2}} - \frac{1}{T_{1}}]

= positive

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B

Δ H

= positive

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C

l o g K_{2} > l o g K_{1}

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D

K_{2} > K_{1}

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Solution

The correct option is B $[\frac{1}{T_{2}} - \frac{1}{T_{1}}]$ = positive
Given that,
$l o g \frac{K_{2}}{K_{1}} = \frac{- △ H}{2.303} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

For endothermic reaction, $△ H > 0$

Now, for $T_{2} > T_{1}$

$\Rightarrow l o g (\frac{K_{2}}{K_{1}}) = \frac{△ H}{2.303} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

$\Rightarrow l o g (\frac{K_{2}}{K_{1}}) > 0 \Rightarrow l o g K_{2} - l o g K_{1} > 0$
$\Rightarrow l o g K_{2} > l o g K_{1}$
$\Rightarrow K_{2} > K_{1}$

$[\frac{1}{T_{2}} - \frac{1}{T_{1}}] < 0$ since $T_{2} > T_{1}$ .

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Similar questions

Q. Effect of increasing temperature on equilibrium constant is given by

l o g K_{2} - l o g K_{1} = \frac{- Δ H}{2.303 R}

[\frac{1}{T_{2}} - \frac{1}{T_{1}}]

. Then for an endothermic reaction the false statement is:

Q. For a reversible reaction

a A + b B ⇌ c C + d D

; the variation of

K

with temperature is given by

log \frac{K_{2}}{K_{1}} = \frac{- Δ H^{o}}{2.303 R} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]

K_{1}

K_{2}

are the equilibrium constants at temperature

T_{1}

T_{2}

T_{2} > T_{1}

Q. Match the column I with column II and mark the appropriate choice.

Column I	Column II
(A) Zero order	(i) $log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$
(B) First order	(ii) $Δ H > 0$
(C) Endothermic reaction	(iii) $k = \frac{2.303}{t} log \frac{[A]_{0}}{[A]}$
(D) Activation energy	(iv) $k = \frac{1}{t} ([A]_{0} - [A])$

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

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

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