Question
Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant
At 500 K, for a reversible reaction A2(g)+B2(g)⇌2AB(g) in a closed container, KC = 2 × 10-5. In the presence of catalyst, the equilibrium is attaining 10 times faster. The equilibrium constant KC in the presence of catalyst at the same temperature is
A
2 × 10-4
B
2 × 10-6
C
2 × 10-10
D
2 × 10-5
Answer
2 × 10-5
Explanation
Solution
The correct answer is (D) : 2 × 10-5.