For a given exothermic reaction K _ p and K' _ p are the equilibrium constants at temperatures T _ 1 and T _ 2 respectively (T 2 .>T 1 ). Assuming that heat of reaction is constant in temperatures range between T _ 1

Correct answer: A. K _ p > K' _ p log K _ 2 K _ 1 = ∆ H ^ o 2.303R ( 1 T _ 1 - 1 T _ 2 ) T _ 2 > T _ 1 So K _ p > K' _ p (exothermic reaction) (Since T _ 2 > T _ 1 )

NEET 2014

Exam: NEET
Year: 2014
Subject: Chemistry
Chapter: Chemical Equilibrium

NEET 2014 · Chemistry · Chemical Equilibrium

For a given exothermic reaction K _ p and K' _ p are the...

For a given exothermic reaction

{K}_{p}

and

{K'}_{p}

are the equilibrium constants at temperatures

{T}_{1}

and

{T}_{2}

respectively (T₂.>T₁ ). Assuming that heat of reaction is constant in temperatures range between

{T}_{1}

and

{T}_{2},

it is readily observed that:

Options

A ${K}_{p} > {K'}_{p}$
B ${K}_{p} < {K'}_{p}$
C ${K}_{p} = {K'}_{p}$
D ${K}_{p} = \dfrac{1}{{K'}_{p}}$

Correct answer

A. ${K}_{p} > {K'}_{p}$

Step-by-step solution

$log \dfrac{{K}_{2}}{{K}_{1}} = \dfrac{∆ {H}^{o}}{2.303R}\left(\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}}\right)$
${T}_{2} > {T}_{1}$ So ${K}_{p} > {K'}_{p}$ (exothermic reaction)
(Since ${T}_{2} > {T}_{1}$ )

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