Equilibrium H2S Db9881

1. **State the problem:** We have the reaction $$2 \text{H}_2\text{O}(g) + 2 \text{SO}_2(g) \rightleftharpoons 2 \text{H}_2\text{S}(g) + 3 \text{O}_2(g)$$ with initial concentrations $[\text{H}_2\text{O}]_0 = 2.80$ M and $[\text{SO}_2]_0 = 2.60$ M. We want to find the equilibrium concentration of $\text{H}_2\text{S}$ given $K_c = 1.3 \times 10^{-6}$. 2. **Write the expression for the equilibrium constant $K_c$: ** $$K_c = \frac{[\text{H}_2\text{S}]^2 [\text{O}_2]^3}{[\text{H}_2\text{O}]^2 [\text{SO}_2]^2}$$ 3. **Define the change in concentration:** Let $x$ be the change in moles per liter of $\text{H}_2\text{S}$ formed at equilibrium. - Since $2$ moles of $\text{H}_2\text{S}$ are formed, the increase in $\text{H}_2\text{S}$ concentration is $+2x$. - For every $2$ moles of $\text{H}_2\text{S}$ formed, $2$ moles of $\text{H}_2\text{O}$ and $2$ moles of $\text{SO}_2$ are consumed, so their concentrations decrease by $2x$ each. - $\text{O}_2$ is produced at $3$ moles per $2$ moles of $\text{H}_2\text{S}$, so its concentration increases by $3x$. 4. **Write equilibrium concentrations:** $$[\text{H}_2\text{O}] = 2.80 - 2x$$ $$[\text{SO}_2] = 2.60 - 2x$$ $$[\text{H}_2\text{S}] = 0 + 2x = 2x$$ $$[\text{O}_2] = 0 + 3x = 3x$$ 5. **Substitute into $K_c$ expression:** $$K_c = \frac{(2x)^2 (3x)^3}{(2.80 - 2x)^2 (2.60 - 2x)^2} = 1.3 \times 10^{-6}$$ 6. **Simplify numerator:** $$(2x)^2 = 4x^2$$ $$(3x)^3 = 27x^3$$ So numerator is $$4x^2 \times 27x^3 = 108 x^5$$ 7. **Assuming $x$ is small compared to initial concentrations, approximate:** $$(2.80 - 2x) \approx 2.80$$ $$(2.60 - 2x) \approx 2.60$$ 8. **Calculate denominator:** $$(2.80)^2 (2.60)^2 = (7.84)(6.76) = 53.0384$$ 9. **Set up approximate equation:** $$\frac{108 x^5}{53.0384} = 1.3 \times 10^{-6}$$ 10. **Solve for $x^5$:** $$x^5 = \frac{1.3 \times 10^{-6} \times 53.0384}{108} = \frac{6.895 \times 10^{-5}}{108} = 6.39 \times 10^{-7}$$ 11. **Take fifth root:** $$x = (6.39 \times 10^{-7})^{1/5}$$ Calculate: $$\log_{10}(6.39 \times 10^{-7}) = \log_{10}(6.39) - 7 = 0.805 - 7 = -6.195$$ Divide by 5: $$-6.195 / 5 = -1.239$$ So: $$x = 10^{-1.239} = 0.0576$$ 12. **Calculate equilibrium concentration of $\text{H}_2\text{S}$:** $$[\text{H}_2\text{S}] = 2x = 2 \times 0.0576 = 0.1152 \text{ M}$$ 13. **Check assumption:** $2x = 0.1152$ is much smaller than initial concentrations $2.80$ and $2.60$, so approximation is valid. **Final answer:** $$\boxed{[\text{H}_2\text{S}]_{eq} = 0.115 \text{ M}}$$