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

left and equilibrium constant decreases

left and equilibrium constant increases

right and equilibrium constant increases

right and equilibrium constant decreases

Correct answer is A

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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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

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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

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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