For the reaction , the equilibrium constant is at a specific temperature. If a mixture is artificial — Chemical Equilibrium Chemistry Question
Question
For the reaction $2A(g) + B(g) \rightleftharpoons 3C(g)$, the equilibrium constant $K_c$ is $4.0$ at a specific temperature. If a mixture is artificially prepared such that $[A] = 0.5\text{ M}$, $[B] = 2.0\text{ M}$, and $[C] = 1.0\text{ M}$, in which direction will the reaction predominantly proceed and why?
Answer: A
💡 Solution & Explanation
The reaction quotient $Q_c$ is calculated as $Q_c = \frac{[C]^3}{[A]^2[B]} = \frac{(1.0)^3}{(0.5)^2(2.0)} = \frac{1}{0.25 \times 2} = \frac{1}{0.5} = 2.0$. Since $Q_c$ ($2.0$) is less than $K_c$ ($4.0$), the system is not at equilibrium. To reach equilibrium, $Q_c$ must increase, meaning the reaction will proceed in the forward direction.
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