2SO\(_2\) \(_{(g) }\)+ O\(_2\) \(_{(g) }\) ↔ 2SO\(_3\)\(_{(g) }\) ΔH = -395.7kJmol\(^{-1}\) In the equation,...

JAMB 2018

2SO\(_2\) \(_{(g) }\)+ O\(_2\) \(_{(g) }\) ↔ 2SO\(_3\)\(_{(g) }\) ΔH = -395.7kJmol\(^{-1}\)
In the equation, an increase in temperature will shift the equilibrium position to the

A. left and equilibrium constant decreases
B. left and equilibrium constant increases
C. right and equilibrium constant increases
D. right and equilibrium constant decreases

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Correct Answer: Option A

Explanation

This is typical of what happens with any equilibrium where the forward reaction is exothermic. Increasing the temperature decreases the value of the equilibrium constant. Where the forward reaction is endothermic, increasing the temperature increases the value of the equilibrium constant. Endothermic reaction has a Positive Enthalpy change. Exothermic reaction has a negative enthalpy change.

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