Question

A mixture of \(\mathrm{H}_{2}, \mathrm{~S}\), and \(\mathrm{H}_{2} \mathrm{~S}\) is held in a 1.0-L vessel at \(90^{\circ} \mathrm{C}\) until the following equilibrium is achieved: $$ \mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g) $$ At equilibrium the mixture contains \(0.46 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(0.40 \mathrm{~g} \mathrm{H}_{2} .\) (a) Write the equilibrium-constant expression for this reaction. (b) What is the value of \(K_{c}\) for the reaction at this temperature? (c) Why can we ignore the amount of \(S\) when doing the calculation in part (b)?

Short answer

a) The equilibrium-constant expression for this reaction is: \( K_c = \frac{[\mathrm{H_2S}]}{[\mathrm{H_2}]} \) b) The value of Kc for this reaction at this temperature is: 0.0682 c) We can ignore the amount of S because it is a solid substance and does not affect the value of the equilibrium constant.

Step-by-step solution

Write the equilibrium-constant expression for the reaction

In order to write the equilibrium-constant expression, we can use the formula: \[ K_c = \frac{[\text{Products}]^{\text{stoichiometric coefficients}}}{[\text{Reactants}]^{\text{stoichiometric coefficients}}} \] For our given reaction: \[ \mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2}\mathrm{S}(g) \] We can then write the equilibrium-constant expression as: \[ K_c = \frac{[\mathrm{H_2S}]}{[\mathrm{H_2}]} \] Notice that the amount of S is not included in the equilibrium expression since it is a solid.

Calculate the equilibrium concentrations

We're given the masses of H₂S and H₂ at equilibrium: - H₂S: 0.46 g - H₂: 0.40 g We need to convert these masses into moles and then find their concentrations in the 1.0 L vessel. Molar mass of H₂S = 1(2.02 g/mol) + 1(32.06 g/mol) = 34.08 g/mol Moles of H₂S = mass / molar mass = 0.46 g / 34.08 g/mol = 0.0135 mol Concentration of H₂S = moles / volume = 0.0135 mol / 1.0 L = 0.0135 M Molar mass of H₂ = 2(1.01 g/mol) = 2.02 g/mol Moles of H₂ = mass / molar mass = 0.40 g / 2.02 g/mol = 0.198 mol Concentration of H₂ = moles / volume = 0.198 mol / 1.0 L = 0.198 M

Calculate the equilibrium constant, Kc

Now that we have found the equilibrium concentrations of H₂S and H₂, we can use the previously written equilibrium-constant expression to find the value of Kc: \[ K_c = \frac{[\mathrm{H_2S}]}{[\mathrm{H_2}]} =\frac{0.0135}{0.198}= 0.0682 \]

Explain the reason for ignoring the amount of S

In equilibrium calculations, we can ignore the amount of solid substances because their concentrations remain constant and they generally do not affect the value of the equilibrium constant. Solids and pure liquids (like water) are not included in the equilibrium-constant expression, so their amounts are irrelevant to the calculation of Kc. That's why we don't need to consider the amount of S (sulfur) when calculating Kc in this exercise. In conclusion: a) The equilibrium-constant expression for this reaction is: \[ K_c = \frac{[\mathrm{H_2S}]}{[\mathrm{H_2}]} \] b) The value of Kc for this reaction at this temperature is: 0.0682 c) We can ignore the amount of S because it is a solid substance and does not affect the value of the equilibrium constant.

Key concepts

Equilibrium-Constant Expression

Understanding the equilibrium-constant expression is crucial for comprehending how chemical reactions reach a state of balance. The expression illustrates the relationship between the concentrations of reactants and products when a reaction is at equilibrium. It's based on the law of mass action, which states that for a reversible reaction at equilibrium and a constant temperature, a certain ratio of reactant and product concentrations has a constant value, known as the equilibrium constant, denoted by K.

For a general reaction, where reactants A and B form products C and D, with lowercase letters representing the stoichiometric coefficients, the equilibrium-constant expression is written as:

\[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
This equation shows that the concentrations of products are raised to the power of their respective stoichiometric coefficients and multiplied together, forming the numerator. Similarly, the concentrations of reactants form the denominator. It's important that the concentration of each species is in molarity (M), which is moles per liter.

In our specific exercise, the equilibrium expression for the formation of hydrogen sulfide (H2S) from hydrogen gas (H2) and solid sulfur (S) is remarkably simpler because we ignore the concentration of solid sulfur in the equation, as its concentration does not change:

\[ K_c = \frac{[\mathrm{H_2S}]}{[\mathrm{H_2}]} \]
Here, we have directly proportional relationship between the formation of H2S and the consumption of H2. In general, solids and liquids are omitted from the equilibrium expression, focusing solely on the gaseous and aqueous species involved in the reaction.

Equilibrium Calculations

Equilibrium calculations are a systematic approach to determining the concentrations of reactants and products at equilibrium. These calculations are essential to quantify the equilibrium constant, K, and to predict the direction in which a reaction will proceed under given conditions.

The process involves several critical steps: determining molar masses, converting masses to moles, finding molar concentrations, and finally, applying these concentrations to the equilibrium-constant expression.

For instance, in the provided exercise, the equilibrium concentrations were calculated by using the molar mass of each gas:

• Molar mass of H2S = 34.08 g/mol
• Moles of H2S = 0.46 g / 34.08 g/mol = 0.0135 mol
• Concentration of H2S = 0.0135 mol / 1.0 L = 0.0135 M
• Molar mass of H2 = 2.02 g/mol
• Moles of H2 = 0.40 g / 2.02 g/mol = 0.198 mol
• Concentration of H2 = 0.198 mol / 1.0 L = 0.198 M

Once these concentrations are known, they can be plugged into the equilibrium-constant expression to find the value of K, which, for our exercise, resulted in Kc being 0.0682. This calculation helps predict how far a reaction has proceeded towards equilibrium and is a valuable tool for chemists in both industrial and research settings.

H2S Equilibrium Concentration

In a chemical reaction involving gases like the production of hydrogen sulfide (H2S), understanding the equilibrium concentration of each species is vital to understanding the extent of reaction progress. The equilibrium concentration is defined as the concentration of a substance in a reaction mixture when the rates of the forward and reverse reactions are equal, causing no net change in concentration.

During calculations, we pay close attention to the moles of the gases involved because it directly relates to their concentrations at equilibrium. For example, in the exercise involving the reaction between hydrogen gas and solid sulfur to form H2S, we calculated the moles of H2S using its molar mass and then determined its equilibrium concentration to be 0.0135 M in a 1.0 L vessel.

Finding the H2S equilibrium concentration is not merely a step in solving a textbook problem; it provides insight into the practical aspects of a chemical reaction. It allows us to predict how much product can be formed from a given amount of reactants, assess the efficiency of reactions, and, if necessary, optimize conditions to achieve a desired yield of product. This is why a clear understanding of not just the methodology, but also the implications of equilibrium concentrations, is essential for students and practitioners of chemistry alike.