The equilibrium constant for the reaction is at . What is the value of for the reverse reaction ? — Chemical Equilibrium Chemistry Question
Question
The equilibrium constant for the reaction $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$ is $0.014$ at $698\text{ K}$. What is the value of $K$ for the reverse reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$?
Answer: B
💡 Solution & Explanation
Reversing a chemical equation results in an equilibrium constant that is the reciprocal of the original. $K_{reverse} = 1 / 0.014 = 71.4$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes