Question

Hydrogen sulfide decomposes according to the following reaction, for which \(K_{c}=9.30 \times 10^{-8}\) at \(700^{\circ} \mathrm{C}:\) $$ 2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) $$ If \(0.45 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{~S}\) is placed in a 3.0 - \(\mathrm{L}\) container, what is the equilibrium concentration of \(\mathrm{H}_{2}(g)\) at \(700^{\circ} \mathrm{C} ?\)

Short answer

1.62 × 10^{-3} M

Step-by-step solution

Write the equation for the equilibrium constant

Given the reaction \[ 2 \text{H}_2 \text{S}(g) \rightleftharpoons 2 \text{H}_2(g) + \text{S}_2(g) \] and the equilibrium constant expression \[ K_c = \frac{[\text{H}_2]^2 [\text{S}_2]}{[\text{H}_2 \text{S}]^2}, \] where \( K_c = 9.30 \times 10^{-8} \).

Initial concentrations

Calculate the initial concentration of \( \text{H}_2 \text{S} \), using the formula \[ [\text{H}_2 \text{S}]_0 = \frac{0.45 \text{ mol}}{3.0 \text{ L}} = 0.15 \text{ M}. \]

Define changes in concentrations

Let \( x \) be the change in concentration of \( \text{H}_2(g) \) and \( \text{S}_2(g) \) at equilibrium. Therefore, the changes are:\[ [\text{H}_2 \text{S}] = 0.15 - 2x, [\text{H}_2] = 2x, [\text{S}_2] = x. \]

Substitute into the equilibrium expression

Substitute the equilibrium concentrations into the equilibrium constant expression:\[ K_c = \frac{(2x)^2 (x)}{(0.15 - 2x)^2} = 9.30 \times 10^{-8}. \]

Simplify and solve the cubic equation

Solve the cubic equation. As this is simplified, assume \( 0.15 - 2x \approx 0.15 \) (since \( K_c \) is very small):\[ K_c \approx \frac{(2x)^2 x}{0.15^2}. \] Solve for \( x \): \[ 9.30 \times 10^{-8} \approx \frac{4x^3}{0.0225}. \] \[ x^3 \approx \frac{9.30 \times 10^{-8} \times 0.0225}{4}. \] \[ x^3 \approx 5.24 \times 10^{-10}. \] \[ x \approx 8.08 \times 10^{-4} \text{ M}. \]

Calculate equilibrium concentration of \( \text{H}_2(g) \)

Since \( [\text{H}_2] = 2x \), substitute \( x \) back in: \[ [\text{H}_2] = 2 \times 8.08 \times 10^{-4} = 1.62 \times 10^{-3} \text{ M}. \]

Key concepts

Equilibrium Constant

In chemical equilibrium, the equilibrium constant (\text{K}_c) is a crucial value that helps in understanding the ratio of product concentrations to reactant concentrations at equilibrium. It is specific to a given reaction at a specific temperature. For the decomposition of hydrogen sulfide (\text{H}_2\text{S}):

$$2 \text{H}_2 \text{S} (g) \rightleftharpoons 2 \text{H}_2(g) + \text{S}_2(g),$$

The expression for the equilibrium constant (\text{K}_c) is given by:

$$K_c = \frac{[\text{H}_2]^2[\text{S}_2]}{[\text{H}_2\text{S}]^2}. $$

Here, \text{K}_c = 9.30 \times 10^{-8} at 700°C. This small value of \text{K}_c signifies that at equilibrium, the concentration of the reactants (\text{H}_2\text{S}) is much higher than the concentration of the products (\text{H}_2 and \text{S}_2).

Concentration Calculation

Calculating concentrations involves using the initial conditions and changes that occur as the system reaches equilibrium. To find the equilibrium concentration of hydrogen gas (\text{H}_2) at 700°C, you can start by determining the initial concentration of \text{H}_2\text{S}. Given:

• 0.45 mol of \text{H}_2\text{S}
• Volume = 3.0 L

The initial concentration is calculated as: $$[\text{H}_2\text{S}]_0 = \frac{0.45 \text{ mol}}{3.0 \text{ L}} = 0.15 \text{ M}.$$

At equilibrium, let

Le Chatelier's Principle

Le Chatelier's Principle is fundamental in predicting how an equilibrium will shift when subjected to a change in conditions. If the system at equilibrium is disturbed by a change in concentration, pressure, volume, or temperature, the equilibrium will shift in a direction that counteracts the disturbance, re-establishing equilibrium. For our reaction:

$$2 \text{H}_2 \text{S} (g) \rightleftharpoons 2 \text{H}_2(g) + \text{S}_2(g),$$

Increasing the concentration of \text{H}_2\text{S} will shift the equilibrium to the right, producing more \text{H}_2 and \text{S}_2. Conversely, increasing the concentration of \text{H}_2 or \text{S}_2 will shift the equilibrium to the left, forming more \text{H}_2\text{S}.

Reaction Quotient

The Reaction Quotient (Q) provides a snapshot of the concentrations of reactants and products at any point in time, not just at equilibrium. For the reaction:

$$2 \text{H}_2 \text{S}(g) \rightleftharpoons 2 \text{H}_2(g) + \text{S}_2(g),$$

• The expression for Q is similar to \text{K}_c:
• Q = \frac{[\text{H}_2]^2[\text{S}_2]}{[\text{H}_2\text{S}]^2}.

Comparing Q to \text{K}_c helps predict the direction in which the reaction will proceed:

• If Q < \text{K}_c, the reaction moves forward, producing more products.
• If Q > \text{K}_c, the reaction goes in reverse, producing more reactants.

For your calculations, you can use the initial concentrations to find Q and understand whether the system is in equilibrium or needs adjustment.