Question

At \(1130^{\circ} \mathrm{C}\), the equilibrium constant \(\left(K_{\mathrm{c}}\right.\) ) for the reaction: $$2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftarrows 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g)$$ is \(2.25 \times 10^{-4} .\) If \(\left[\mathrm{H}_{2} \mathrm{~S}\right]=4.84 \times 10^{-3} \mathrm{M}\) and $$\left[\mathrm{H}_{2}\right]=1.50 \times 10^{-3} M, \text { calculate }\left[\mathrm{S}_{2}\right]$$.

Short answer

\([\mathrm{S}_{2}] = 2.34 \times 10^{-3}\, M\).

Step-by-step solution

Write the expression for the equilibrium constant

The equilibrium constant expression for the reaction \(2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftarrows 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g)\) is given by the ratio of the concentration of products to reactants, each raised to the power of their coefficients in the balanced equation: \[K_c = \frac{[\mathrm{H}_{2}]^2 [\mathrm{S}_{2}]}{[\mathrm{H}_{2} \mathrm{~S}]^2}\]Given that \(K_c = 2.25 \times 10^{-4}\), \([\mathrm{H}_{2} \mathrm{~S}] = 4.84 \times 10^{-3}\, M\), and \([\mathrm{H}_{2}] = 1.50 \times 10^{-3}\, M\), we need to find \([\mathrm{S}_{2}]\).

Plug known values into the equilibrium expression

Substitute the known values of \(K_c\), \([\mathrm{H}_{2}]\), and \([\mathrm{H}_{2} \mathrm{~S}]\) into the equilibrium constant expression:\[2.25 \times 10^{-4} = \frac{(1.50 \times 10^{-3})^2 [\mathrm{S}_{2}]}{(4.84 \times 10^{-3})^2}\]

Solve for \([\mathrm{S}_{2}]\)

Rearrange the equation to solve for \([\mathrm{S}_{2}]\):\[[\mathrm{S}_{2}] = \frac{(2.25 \times 10^{-4})(4.84 \times 10^{-3})^2}{(1.50 \times 10^{-3})^2}\]Calculate the values inside the equation:- Calculate \((1.50 \times 10^{-3})^2 = 2.25 \times 10^{-6}\)- Calculate \((4.84 \times 10^{-3})^2 = 2.34 \times 10^{-5}\)Then, substitute back:\[[\mathrm{S}_{2}] = \frac{(2.25 \times 10^{-4})(2.34 \times 10^{-5})}{2.25 \times 10^{-6}}\]

Final Calculation and Result

Complete the calculation to find \([\mathrm{S}_{2}]\):\[[\mathrm{S}_{2}] = \frac{2.25 \times 10^{-4} \times 2.34 \times 10^{-5}}{2.25 \times 10^{-6}} = \frac{5.265 \times 10^{-9}}{2.25 \times 10^{-6}}\]Divide to get:\[[\mathrm{S}_{2}] = 2.34 \times 10^{-3}\, M\]

Verify Units and Calculation

Verify the units match. Here all concentrations are in molarity \(M\). Double-check calculations to ensure they were performed accurately. Each component part follows the equilibrium setup correctly to reach the final concentration of \([\mathrm{S}_{2}].\)

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs when a reversible chemical reaction ceases to change its ratio of reactants and products. This state is achieved when the rates of the forward and reverse reactions are equal. At equilibrium, the concentrations of reactants and products remain constant over time. This doesn't mean they are equal, just stable in their respective ratios.

• Equilibrium is dynamic; reactions continue to occur, but no net change in concentrations happens.
• In our exercise, equilibrium concerns the reaction: \[2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftarrows 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g)\]
• The equilibrium constant \(K_c\) quantitatively expresses the ratio of the concentrations of products to reactants at equilibrium, for a given temperature.

Understanding chemical equilibrium is crucial to predicting the outcome of reactions and manipulating them for desired products.

Reaction Quotient

The reaction quotient \(Q\) is similar to the equilibrium constant \(K_c\), but it applies to reactions that are not at equilibrium. \(Q\) is calculated using the same formula as \(K_c\) but based on the initial concentrations rather than those at equilibrium.

• If \(Q = K_c\), the reaction is at equilibrium.
• If \(Q < K_c\), the reaction will proceed forward to produce more products.
• If \(Q > K_c\), the reaction will shift in the reverse direction to form more reactants.

Understanding the reaction quotient helps predict the direction in which a reaction will proceed to reach equilibrium.

Molarity

Molarity, denoted as \(M\), is the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. This measure is crucial in understanding reaction dynamics, especially when calculating equilibrium constants.

• Molarity allows chemists to express and calculate concentrations across different solutions.
• In the original problem, concentrations like \([\mathrm{H}_{2} \mathrm{S}]=4.84\times 10^{-3} \mathrm{M}\) and \([\mathrm{H}_{2}]=1.50 \times 10^{-3} \mathrm{M}\) are given in molarity.
• Correctly employing molarity is essential for using equilibrium expressions accurately.

Without consistent units of concentration like molarity, comparing and computing reactions accurately wouldn't be possible.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is based on the conservation of mass where the number of atoms for each element in the reactants' side equals that on the products' side.

• The coefficients in a balanced chemical equation tell you the ratios of moles of each substance involved in the reaction.
• In our equilibrium expression \(K_c = \frac{[\mathrm{H}_{2}]^2 [\mathrm{S}_{2}]}{[\mathrm{H}_{2} \mathrm{~S}]^2}\), these coefficients (like the '2' in \(2 \mathrm{H}_{2} \mathrm{S}\)) become exponents when calculating \(K_c\).
• Accurate stoichiometric ratios are critical in determining correct equilibrium expressions.

Grasping stoichiometry ensures precise measurements and outcomes in chemical reactions and related calculations.