[Single-digit Integer] The per cent dissociation of (g) if 0.1 mole of is kept in 0.4 L vessel at 10 — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate the Initial Concentration of Reactant \[C = \frac{n}{V} = \frac{0.1\text{ mol}}{0.4\text{ L}} = 0.25\text{ M}\] Step 2 - Construct the ICE Table The dissociation: \[\ce{2H2S(g) <=> 2H2(g) + S2(g)}\] Let $\alpha$ = degree of dissociation: \[\begin{array}{lccccc} \text{Species} & \ce{2H2S(g)} & \ce{<=>} & \ce{2H2(g)} & + & \ce{S2(g)} \ \hline \text{Initial (M)} & C & & 0 & & 0 \ \text{Change (M)} & -C\alpha & & +C\alpha & & +\frac{C\alpha}{2} \ \text{Equil. (M)} & C(1-\alpha) & & C\alpha & & \frac{C\alpha}{2} \ \end{array}\] Step 3 - Express and Simplify \(K_c\) \[K_c = \frac{[\ce{H2}]^2[\ce{S2}]}{[\ce{H2S}]^2} = \frac{(C\alpha)^2 \cdot \frac{C\alpha}{2}}{[C(1-\alpha)]^2} = \frac{C\alpha^3}{2(1-\alpha)^2}\] Since $K_c = 1.0 \times 10^{-6}$ is very small, $\alpha \ll 1$, so $(1-\alpha) \approx 1$: \[K_c \approx \frac{C\alpha^3}{2}\] Step 4 - Solve for $\alpha$ and Percent Dissociation \[1.0 \times 10^{-6} = \frac{0.25 \times \alpha^3}{2}\] \[\alpha^3 = \frac{2.0 \times 10^{-6}}{0.25} = 8.0 \times 10^{-6}\] \[\alpha = (8.0 \times 10^{-6})^{1/3} = 2.0 \times 10^{-2} = 0.02\] \[\text{Percent Dissociation} = \alpha \times 100\% = 0.02 \times 100\% = \boxed{2\%}\]