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 and T 2 , it is readily observed that:NEET 2014

log ⁡ K 2 K 1 = ∆ H o 2.303 R 1 T 1 - 1 T 2 T 2 > T 1 So K p > K ' p (exothermic reaction) (Since T 2 > T 1 )

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Chemistry - Chemical Equilibrium Question with Solution | TestHub

ChemistryChemical EquilibriumLe-Chatelier's PrincipleHard2 minPYQ_2014

ChemistryHardsingle choice

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:NEET 2014

Options:

Answer:

Solution:

$\log \frac{K_{2}}{K_{1}} = \frac{∆ H^{o}}{2.303 R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$
$T_{2} > T_{1}$ So $K_{p} > {K'}_{p}$ (exothermic reaction)
(Since $T_{2} > T_{1}$ )

Stream:NEETSubject:ChemistryTopic:Chemical EquilibriumSubtopic:Le-Chatelier's Principle

⏱ 2mℹ️ Source: PYQ_2014

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