For the reaction $2H_2S(g) = 2H_2(g) + S_2(g)$, the equilibrium constant $K_c = 4.2 \times 10^{-6}$ at 830°C. What are the equilibrium concentrations when 0.500 mol $H_2S$ is place... For the reaction $2H_2S(g) = 2H_2(g) + S_2(g)$, the equilibrium constant $K_c = 4.2 \times 10^{-6}$ at 830°C. What are the equilibrium concentrations when 0.500 mol $H_2S$ is placed in an empty 1.0 L vessel at 830°C? What fraction of the $H_2S$ dissociated? Read more

Steps to Solve

1. Write out the balanced chemical equation

The balanced chemical equation is given as: $2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g)$

2. Set up an ICE table

Set up an ICE (Initial, Change, Equilibrium) table to track the concentrations of the reactants and products. Since the volume is 1.0 L, the number of moles is equal to the molar concentration.

3. Write the expression for the equilibrium constant $K_c$

The equilibrium constant $K_c$ is given by: $K_c = \frac{[H_2]^2[S_2]}{[H_2S]^2}$

4. Substitute the equilibrium concentrations from the ICE table into the $K_c$ expression

Substitute the equilibrium concentrations from the ICE table into the $K_c$ expression: $4.2 \times 10^{-6} = \frac{(2x)^2(x)}{(0.500 - 2x)^2}$

5. Solve for x by making an assumption to simplify the equation

Since $K_c$ is very small, we can assume that $2x$ is much smaller than 0.500, so $0.500 - 2x \approx 0.500$. This simplifies the equation to: $4.2 \times 10^{-6} = \frac{4x^3}{(0.500)^2}$ $4x^3 = 4.2 \times 10^{-6} \times (0.500)^2$ $4x^3 = 1.05 \times 10^{-6}$ $x^3 = 0.2625 \times 10^{-6}$ $x = \sqrt[3]{0.2625 \times 10^{-6}}$ $x = 6.40 \times 10^{-3}$

6. Check the assumption

Check if the assumption $2x << 0.500$ is valid: $2x = 2 \times 6.40 \times 10^{-3} = 0.0128$ $\frac{0.0128}{0.500} \times 100% = 2.56%$ Since 2.56% is less than 5%, the assumption is valid.

7. Calculate the equilibrium concentrations

Now, calculate the equilibrium concentrations: $[H_2S] = 0.500 - 2x = 0.500 - 2(6.40 \times 10^{-3}) = 0.500 - 0.0128 = 0.487 \ M$ $[H_2] = 2x = 2(6.40 \times 10^{-3}) = 0.0128 \ M$ $[S_2] = x = 6.40 \times 10^{-3} \ M$

8. Calculate the fraction of $H_2S$ dissociated

The fraction of $H_2S$ dissociated is the amount of $H_2S$ that reacted ($2x$) divided by the initial amount of $H_2S$ (0.500 M): Fraction dissociated $= \frac{2x}{0.500} = \frac{0.0128}{0.500} = 0.0256$