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Relation between Kp and Kc

Effect of inc...

Question

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:

A

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

positive

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B

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

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C

Δ H =

positive

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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, $l o g \frac{K_{2}}{K_{1}} = \frac{- △ H}{2.303 R} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$
(i) For Endothermic process, heat is required. Hence $△ H = p o s i t i v e$
(ii) Also, $T_{2} > T_{1}$ , i.e $T_{2}$ is at higher temperature than $T_{1}$
$\Rightarrow (\frac{1}{T_{2}} - \frac{1}{T_{1}})$
$(\frac{T_{1} - T_{2}}{T_{1} T_{2}}) < 0$
$T_{1} - T_{2} < 0$
(iii) $l o g \frac{K_{2}}{K_{1}} = \frac{- △ H}{2.303 R} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$
$l o g \frac{K_{2}}{K_{1}} > 0$
$l o g K_{2} > l o g K_{1}$

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Similar questions

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:

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} a n d K_{3}

are equilibrium constant for formation of

A D, A D_{2} a n d A D_{3}

respectively as follow

A + D ⇌ A D, A D + D ⇌ A D_{2}, A D_{2} + D ⇌ A D_{3}

then equilibrium constant 'K" for

A + 3 D ⇌ A D_{3}

Q. Vant' Hoff equation shows the effect of temperature on equilibrium constants

K_{c}

K_{p} .

K_{p}

varies with temperature according to the relation:

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) $k = \frac{1}{t} ([A]_{0} - [A])$

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