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Le Chatelier's Principle for Del N Equals to 0

For a reversi...

Question

For a reversible reaction $a A + b B ⇌ c C + d D$ ; the variation of K with temperature is given by log
$\frac{K_{2}}{K_{1}} = \frac{- Δ H^{o}}{2.303 R} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$ then,

A

K_{2} > K_{1}

if

T_{2} > T_{1}

for an endothermic change

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B

K_{2} < K_{1}

if

T_{2} > T_{1}

for an endothermic change

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C

K_{2} > K_{1}

if

T_{2} > T_{1}

for an exothermic change

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D

K_{2} < K_{1}

if

T_{2} > T_{1}

for an exothermic change

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Solution

The correct options are
A $K_{2} > K_{1}$ if $T_{2} > T_{1}$ for an endothermic change
D $K_{2} < K_{1}$ if $T_{2} > T_{1}$ for an exothermic change
$\frac{K_{2}}{K_{1}} = \frac{Δ H^{o}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
$Δ H^{o} (+ v e) \Rightarrow T_{2} > T_{1} \Rightarrow$ Forward reaction $K_{2} > K_{1}$
$Δ H^{o} (- v e) \Rightarrow T_{2} < T_{1} \Rightarrow$ Backward reaction $K_{2} < K_{1}$

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Similar questions

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}}]

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

K_{1}

K_{2}

are the equilibrium constants at temperature

T_{1}

T_{2}

T_{2} > T_{1}

Q. The activation energy for forward and backward reactions are

50 k J / m o l

40 k J / m o l

respectively. If

K_{1}

K_{2}

are the equilibrium constants of reaction at temperature

T_{1}

T_{2}

respectively and

T_{2} > T_{1}

Q. 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:

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