Questions: Hydrogen sulfide decomposes according to the following reaction, for which Kc=9.30 x 10^-8 at 700°C: 2 H2S(g) ⇌ 2 H2(g) + S2(g) If 0.59 mol of H2S is placed in a 3.0-L container, what is the equilibrium concentration of H2(g) at 700°C?

Hydrogen sulfide decomposes according to the following reaction, for which
Kc=9.30 x 10^-8 at 700°C:
2 H2S(g) ⇌ 2 H2(g) + S2(g)

If 0.59 mol of H2S is placed in a 3.0-L container, what is the equilibrium concentration of H2(g) at 700°C?

Transcript text: Hydrogen sulfide decomposes according to the following reaction, for which \[ K_{\mathrm{c}}=9.30 \times 10^{-8} \text { at } 700^{\circ} \mathrm{C}: \] \[ 2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \] If 0.59 mol of $\mathrm{H}_{2} \mathrm{~S}$ is placed in a $3.0-\mathrm{L}$ container, what is the equilibrium concentration of $\mathrm{H}_{2}(\mathrm{~g})$ at $700^{\circ}$ C?

Solution

Solution Steps

Step 1: Write the Expression for the Equilibrium Constant

The equilibrium constant expression for the reaction is given by:

\[ K_{\mathrm{c}} = \frac{[\mathrm{H}_2]^2 [\mathrm{S}_2]}{[\mathrm{H}_2\mathrm{S}]^2} \]

Step 2: Determine Initial Concentrations

Calculate the initial concentration of $\mathrm{H}_2\mathrm{S}$:

\[ [\mathrm{H}_2\mathrm{S}]_0 = \frac{0.59 \, \text{mol}}{3.0 \, \text{L}} = 0.197 \, \text{M} \]

Step 3: Set Up the ICE Table

Define the change in concentration at equilibrium as $x$:

Initial concentrations: $[\mathrm{H}_2\mathrm{S}] = 0.197 \, \text{M}$, $[\mathrm{H}_2] = 0$, $[\mathrm{S}_2] = 0$
Change in concentrations: $[\mathrm{H}_2\mathrm{S}] = -2x$, $[\mathrm{H}_2] = +2x$, $[\mathrm{S}_2] = +x$
Equilibrium concentrations: $[\mathrm{H}_2\mathrm{S}] = 0.197 - 2x$, $[\mathrm{H}_2] = 2x$, $[\mathrm{S}_2] = x$

Step 4: Substitute into the Equilibrium Expression

Substitute the equilibrium concentrations into the equilibrium constant expression:

\[ K_{\mathrm{c}} = \frac{(2x)^2 (x)}{(0.197 - 2x)^2} = 9.30 \times 10^{-8} \]

Step 5: Simplify and Solve for $x$

Assume $x$ is small compared to 0.197, so $0.197 - 2x \approx 0.197$. Substitute and solve for $x$:

\[ \frac{4x^3}{0.197^2} = 9.30 \times 10^{-8} \]

\[ 4x^3 = 9.30 \times 10^{-8} \times 0.197^2 \]

\[ x^3 = \frac{9.30 \times 10^{-8} \times 0.197^2}{4} \]

Calculate $x$ and then find $2x$ for $[\mathrm{H}_2]$.

Step 6: Calculate Equilibrium Concentration of $\mathrm{H}_2$

Calculate $2x$ to find the equilibrium concentration of $\mathrm{H}_2$:

\[ [\mathrm{H}_2] = 2x \]