For the reaction , the equilibrium constant is at a specific temperature. If a mixture is artificially prepared such that , , and , in which direction will the reaction predominantly proceed and why?

Question

For the reaction $2A(g) + B(g) ightleftharpoons 3C(g)$, the equilibrium constant $K_c$ is $4.0$ at a specific temperature. If a mixture is artificially prepared such that $[A] = 0.5	ext{ M}$, $[B] = 2.0	ext{ M}$, and $[C] = 1.0	ext{ M}$, in which direction will the reaction predominantly proceed and why?

Prashant Kotian · Accepted Answer

Correct answer: A. The reaction quotient is calculated as . Since () is less than (), the system is not at equilibrium. To reach equilibrium, must increase, meaning the reaction will proceed in the forward direction.

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