The equilibrium mixture for the reaction: 2(g) ⇌ 2(g) + (g) has 1 mole , 0.20 mole and 0.80 mole of — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Determine the Equilibrium Concentrations The molar concentration of any chemical species in a closed mixture is defined as the ratio of its number of moles to the total volume of the reaction vessel: \[[\text{C}] = \frac{n}{V}\] We are given the following equilibrium quantities in a $V = 2\text{ L}$ vessel: - Moles of hydrogen sulfide, $n_{\ce{H2S}} = 1\text{ mol}$ - Moles of hydrogen, $n_{\ce{H2}} = 0.20\text{ mol}$ - Moles of diatomic sulfur, $n_{\ce{S2}} = 0.80\text{ mol}$ Substituting these values, we calculate the equilibrium concentration for each substance: - Concentration of $\ce{H2S}$: \[[\ce{H2S}] = \frac{1\text{ mol}}{2\text{ L}} = 0.50\text{ M}\] - Concentration of $\ce{H2}$: \[[\ce{H2}] = \frac{0.20\text{ mol}}{2\text{ L}} = 0.10\text{ M}\] - Concentration of $\ce{S2}$: \[[\ce{S2}] = \frac{0.80\text{ mol}}{2\text{ L}} = 0.40\text{ M}\] Step 2 - State the Expression for the Equilibrium Constant ($K_c$) The balanced chemical equation representing the decomposition of hydrogen sulfide gas is: \[\ce{2H2S(g) <=> 2H2(g) + S2(g)}\] According to the law of chemical equilibrium, the concentration-based equilibrium constant ($K_c$) is written as: \[K_c = \frac{[\ce{H2}]^2 [\ce{S2}]}{[\ce{H2S}]^2}\] Step 3 - Substitute Molar Concentrations and Calculate $K_c$ Substituting the equilibrium concentrations calculated in Step 1 into our $K_c$ expression: \[K_c = \frac{(0.10\text{ M})^2 \times (0.40\text{ M})}{(0.50\text{ M})^2}\] Let us compute the numerator and denominator separately: - Numerator: \[(0.10)^2 \times 0.40 = 0.01 \times 0.40 = 0.004\text{ M}^3\] - Denominator: \[(0.50)^2 = 0.25\text{ M}^2\] Dividing the numerator by the denominator to obtain $K_c$: \[K_c = \frac{0.004\text{ M}^3}{0.25\text{ M}^2} = 0.016\text{ M}\] Thus, the equilibrium constant of the reaction is: \[K_c = \boxed{0.016\text{ M}}\] Step 4 - Evaluate the Options - **Option (A) $0.16\text{ M}$**: Incorrect. This represents a calculation error, such as failing to square the concentration terms correctly. - **Option (B) $0.008\text{ M}$**: Incorrect. This value is obtained if one forgets to divide the moles of $\ce{S2}$ by the volume of $2\text{ L}$ while dividing the others. - **Option (C) $0.016\text{ M}$**: Correct. This is the exact value calculated using the correct equilibrium concentrations in the $K_c$ expression. - **Option (D) $0.032\text{ M}$**: Incorrect. This is the value obtained if the initial number of moles is substituted directly without dividing by the volume of $2\text{ L}$: $\frac{(0.20)^2 \times 0.80}{1^2} = 0.032\text{ M}$.