Problem 87 - Chemistry The Science in Context

Question

The value of \(K_{\mathrm{c}}\) for the thermal decomposition of hydrogen sulfide $$ 2 \mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) $$ is \(2.2 \times 10^{-4}\) at \(1400 \mathrm{K} .\) A sample of gas in which \(\left[\mathrm{H}_{2} \mathrm{S}\right]\) \(=6.00 M\) is heated to \(1400 \mathrm{K}\) in a sealed high-pressure vessel. After chemical equilibrium has been achieved, what is the value of \(\left[\mathrm{H}_{2} \mathrm{S}\right]\) ? Assume that no \(\mathrm{H}_{2}\) or \(\mathrm{S}_{2}\) was present in the original sample.

Step-by-Step Solution

Answer

Answer: The equilibrium concentration of H2S is approximately 5.33 M.

Step 1: Write the balanced chemical equation

The given balanced chemical equation is: $$ 2 \mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g) + \mathrm{S}_{2}(g) $$

Step 2: Set up the ICE table

We will create an ICE table to keep track of the initial concentrations, the change in concentrations, and the equilibrium concentrations of the reactants and products. $$ \begin{array}{c|ccc} & [\mathrm{H}_2\mathrm{S}] & [\mathrm{H}_2] & [\mathrm{S}_2] \\ \hline \text {Initial} & 6.00 & 0 & 0 \\ \text {Change} & -2x & +2x & +x \\ \text {Equilibrium} & 6.00 - 2x & 2x & x \end{array} $$ Here, x represents the change in concentration of H2S, H2, and S2.

Step 3: Write the expression for the equilibrium constant \(K_c\)

According to the balanced chemical equation, $$ K_{c} = \frac{[\mathrm{H}_2]^2[\mathrm{S}_2]}{[\mathrm{H}_2\mathrm{S}]^2} $$

Step 4: Substitute known values and solve for x

We are given that \(K_c = 2.2 \times 10^{-4}\), so we can substitute the equilibrium concentrations from the ICE table and solve for x: $$ 2.2 \times 10^{-4} = \frac{(2x)^2(x)}{(6.00 - 2x)^2} $$ Solving the equation for x can be done through algebra, using numerical methods, or using an equation solver tool. The numerical solution for x is approximately 0.334.

Step 5: Calculate the equilibrium concentration of H2S

Use the value of x to find the equilibrium concentration of H2S using the expression from the ICE table: $$ [\mathrm{H}_2\mathrm{S}]_{eq} = 6.00 - 2x = 6.00 - 2(0.334) \approx 5.33 \mathrm{M} $$ The equilibrium concentration of H2S is approximately 5.33 M.

Key Concepts

ICE Table

An ICE table is a great tool to help visualize and manage chemical equilibrium problems. 'ICE' stands for Initial, Change, and Equilibrium. This table helps you calculate how concentrations of reactants and products change as the reaction progresses to equilibrium.

- **Initial Concentrations**: These are the starting amounts of reactants and products before the reaction occurs. In our exercise, the initial concentration of hydrogen sulfide (\( \mathrm{H}_{2}\mathrm{S} \)) is 6.00 M.- **Change in Concentrations**: Represents the increase or decrease in concentrations as the reaction proceeds. The change is typically represented by a variable, often denoted as \( x \). In this case, we assume \( \mathrm{H}_{2} \) and \( \mathrm{S}_{2} \) increase by \(+2x\) and \(+x\) respectively due to the stoichiometry of the reaction.- **Equilibrium Concentration**: The concentrations of reactants and products once the reaction has reached equilibrium. It combines the initial concentration and the change from the reaction progress. For hydrogen sulfide, this would be \(6.00 - 2x\).By setting up an ICE table, it becomes much easier to apply the data to equilibrium expressions and solve for unknowns, such as changes in concentration.

Equilibrium Constant

The equilibrium constant, represented as \( K_c \) in this context because we are dealing with concentrations, is a numerical value that indicates the ratio of product concentrations to reactant concentrations at equilibrium. The equation for a given reaction is derived from the reaction's coefficients.

In our problem, the reaction equation is:\[ 2 \mathrm{H}_{2}\mathrm{S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g) + \mathrm{S}_{2}(g) \]Based on this, the equilibrium constant expression becomes:\[ K_{c} = \frac{[\mathrm{H}_2]^2[\mathrm{S}_2]}{[\mathrm{H}_2\mathrm{S}]^2} \]- **Numerator**: The product concentrations, each raised to the power of its coefficient in the balanced equation. Thus, \([\mathrm{H}_2]^2[\mathrm{S}_2]\).- **Denominator**: The reactant concentrations also raised to the power of their respective coefficients.Given \( K_c = 2.2 \times 10^{-4} \) at 1400 K, this expression defines how proportions of reactants and products balance out in this specific reaction. Solving the equation for unknown values involves inserting known concentrations and solving for those that aren't, using the framework provided by the ICE table.

Thermal Decomposition

Thermal decomposition is a process where a chemical compound breaks down into simpler substances when heated. It's a form of chemical reaction that is often reversible, as is the case with our problem on hydrogen sulfide decomposition.

In the given reaction:\[ 2 \mathrm{H}_{2}\mathrm{S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g) + \mathrm{S}_{2}(g) \]- **Heating**: The sample was heated to \(1400 \mathrm{K}\) which provides sufficient energy for the bonds in hydrogen sulfide molecules to break, leading to the formation of hydrogen and sulfur gas.- **Reversibility**: Since this is an equilibrium reaction, not only does the decomposition of hydrogen sulfide occur, but hydrogen and sulfur can recombine to form hydrogen sulfide again until equilibrium is reached.Understanding the concept of thermal decomposition in equilibrium reactions helps explain why products and reactants might be present simultaneously. This balance between forward decomposition and reverse recombination at high temperatures leads to the dynamic equilibrium observed in our system.