Statement I: For A(g) ⇌ B(g), equilibrium moles of A and B are a and b in 1 L. If 5 moles of A and 3 — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Let $K = b/a$ (given: equilibrium [A] = a mol/L, [B] = b mol/L in 1 L). After adding 5 mol A and 3 mol B to the 1 L system: \[ Q = \frac{[B]_{\text{new}}}{[A]_{\text{new}}} = \frac{b+3}{a+5} \] \textbf{Statement I: "The reaction must move in the forward direction" — INCORRECT} This is only guaranteed if $Q < K$, i.e., $\dfrac{b+3}{a+5} < \dfrac{b}{a}$, which gives $ab + 3a < ab + 5b$, i.e., $3a < 5b$, i.e., $K = b/a > 3/5$. But if $K < 3/5$, then $Q > K$ and the reaction shifts \emph{backward}. Statement I is not universally true. \textbf{Statement II: "The shift depends on how Q compares to K" — CORRECT} The direction of shift is always determined by comparing Q with K regardless of whether more reactant or product was added. Statement II correctly explains this. \textbf{Answer: D} — Statement I is incorrect, Statement II is correct.