For the reaction: X2(g) + Y2(g) ⇌ 2XY(g), 2 moles of 'X2' was taken in a 2 L vessel and 3 moles of ' — Chemical Equilibrium Chemistry Question
Question
For the reaction: X2(g) + Y2(g) ⇌ 2XY(g), 2 moles of 'X2' was taken in a 2 L vessel and 3 moles of 'Y2' was taken in a 3 L vessel. Both vessels were then connected. At equilibrium, concentration of 'XY' is 0.7 M. Equilibrium concentrations of 'X2' and 'Y2' would be:
💡 Solution & Explanation
Reaction: $\text{X}_2(g) + \text{Y}_2(g) \rightleftharpoons 2\text{XY}(g)$ \textbf{Step 1 — Combined initial concentrations:} When the 2 L and 3 L vessels are connected, total volume = 5 L. \[ [\text{X}_2]_0 = \frac{2\ \text{mol}}{5\ \text{L}} = 0.4\ \text{M}, \qquad [\text{Y}_2]_0 = \frac{3\ \text{mol}}{5\ \text{L}} = 0.6\ \text{M} \] \textbf{Step 2 — ICE table (let $x$ mol/L of X$_2$ and Y$_2$ react):} \begin{center} \begin{tabular}{lccc} & $[\text{X}_2]$ & $[\text{Y}_2]$ & $[\text{XY}]$ \\ Initial & 0.4 & 0.6 & 0 \\ Change & $-x$ & $-x$ & $+2x$ \\ Equil. & $0.4-x$ & $0.6-x$ & $2x$ \\ \end{tabular} \end{center} Given: $[\text{XY}]_{\text{eq}} = 0.7\ \text{M} \Rightarrow 2x = 0.7 \Rightarrow x = 0.35$ \textbf{Step 3 — Equilibrium concentrations:} \[ [\text{X}_2]_{\text{eq}} = 0.4 - 0.35 = 0.05\ \text{M} \] \[ [\text{Y}_2]_{\text{eq}} = 0.6 - 0.35 = 0.25\ \text{M} \] \textbf{Answer: D} — 0.05 M and 0.25 M