For the reaction 2NO2(g) in equilibrium with N2O4(g), the enthalpy change is negative 57.2 kJ per mole. What happens to the equilibrium constant when the temperature is increased?

B. It decreases — According to Le Chatelier's principle and the van't Hoff equation, the equilibrium constant for an exothermic reaction decreases with increasing temperature.

JEE Main · ChemistryHard

$For the reaction 2 NO_{2} (g) ⇌ N_{2} O_{4} (g), the enthalpy change is Δ H = - 57.2 kJ/mol . What happens to the equilibrium constant when the temperature is increased?$

A. $It increases$
B. $It decreases$
C. $It remains constant$
D. $It becomes zero$

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Correct answer: B — $It decreases$

Solution

$The reaction is exothermic as Δ H < 0 .$
$The van't Hoff equation states ln (\frac{K _{2}}{K _{1}}) = \frac{- Δ H}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}}) .$
$For an exothermic reaction, increasing temperature (T_{2} > T_{1}) makes the right side negative.$
$Therefore, K_{2} < K_{1} .$
$Hence the answer is (B).$

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For the reaction 2NO2​(g)⇌N2​O4​(g), the enthalpy change is ΔH=−57.2 kJ/mol. What happens to the equilibrium constant when the temperature is increased?

Solution

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More Equilibrium practice questions

$For the reaction 2 NO_{2} (g) ⇌ N_{2} O_{4} (g), the enthalpy change is Δ H = - 57.2 kJ/mol . What happens to the equilibrium constant when the temperature is increased?$