JEE Main Chemistry question
One mole of ethyl alcohol was treated with one mole of acetic acid at 25°C. 2/3 of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be:
Answer: (D)
Solution
Kc=(2/3)²/(1/3)²=4.
One mole of ethyl alcohol was treated with one mole of acetic acid at 25°C. 2/3 of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be:
Kc=(2/3)²/(1/3)²=4.