2H2(g) + O2(g) → 2H2O(g) ΔH = -ve What happens to the equilibrium constant of th...
2H2(g) + O2(g) → 2H2O(g) ΔH = -ve
What happens to the equilibrium constant of the reaction above if the temperature is increased?
Related Lesson: Summarizing Equilibrium Constants | Fundamental Equilibrium Concepts
The correct answer is C.
In the given reaction, 2H2(g) + O2(g) → 2H2O(g), the ΔH is negative, which means that the reaction is exothermic. This means that heat is released as the reaction proceeds.
According to Le Châtelier's principle, if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust itself to counteract the change and restore a new equilibrium. In this case, the change in conditions is an increase in temperature.
As the reaction is exothermic, increasing the temperature shifts the equilibrium position to the left (the side of the reactants) to absorb the excess heat. This means that the forward reaction (formation of water) is slowed down, while the reverse reaction (decomposition of water) is sped up.
As a result, the equilibrium constant (K) of the reaction decreases. This is in line with Option C, which states that the equilibrium constant decreases when the temperature is increased.
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