2SO\(_2\) \(_{(g) }\)+ O\(_2\) \(_{(g) }\) ↔ 2SO\(_3...
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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