Question

A mixture of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) and \(\mathrm{CH}_{4}(\mathrm{g})\) in the mole ratio 2: 1 was brought to equilibrium at \(700^{\circ} \mathrm{C}\) and a total pressure of 1 atm. On analysis, the equilibrium mixture was found to contain \(9.54 \times 10^{-3} \mathrm{mol} \mathrm{H}_{2} \mathrm{S} .\) The \(\mathrm{CS}_{2}\) pre- sent at equilibrium was converted successively to \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and then to \(\mathrm{BaSO}_{4} ; 1.42 \times 10^{-3} \mathrm{mol} \mathrm{BaSO}_{4}\) was obtained. Use these data to determine \(K_{\mathrm{p}}\) at \(700^{\circ} \mathrm{C}\) for the reaction $$\begin{aligned} 2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{CH}_{4}(\mathrm{g}) \rightleftharpoons \mathrm{CS}_{2}(\mathrm{g})+& 4 \mathrm{H}_{2}(\mathrm{g}) \\\ & K_{\mathrm{p}} \text { at } 700^{\circ} \mathrm{C}=? \end{aligned}$$

Short answer

To find the value of the equilibrium constant \(K_p\) at \(700^{\circ}C\), we need to make use of the reported stoichiometric ratio of the reactants, the amount of some of the substances present at equilibrium, and the total pressure. Mathematical manipulation based on these conditions and the reaction equation will lead us to the value of \(K_p\). The exact answer depends on the accurate mathematical calculations.

Step-by-step solution

Calculation of the Initial Moles

Given the reaction 2H2S(g) + CH4(g) ⇌ CS2(g) + 4H2(g). Initially, the mole ratios of \(H_{2}S\) and \(CH_{4}\) are 2:1. Without the exact total moles, assume \(H_{2}S\) has \(2x\) moles and \(CH_{4}\) has \(x\) mole. As a result, once the total initial moles of \(H_{2}S\) and \(CH_{4}\) are determined (as \(3x\)), we will be able to find the actual value of \(x\).

Calculation of the Moles at Equilibrium

Among the molecules in the reaction, only \(H_{2}S\) and \(CS_{2}\) are associated with the amounts at the equilibrium. Therefore, we will use these data to find the value of \(x\). With \(2x\) as the initial moles of \(H_{2}S\), \(2x - 2y\) would be left at the equilibrium, since the reaction uses 2 moles of \(H_{2}S\) to produce a mole of \(CS_{2}\), where \(y\) represents the moles of formed \(CS_{2}\). In this exercise, \(2x - 2y\) equals \(9.54 \times 10^{-3}\), because the moles of \(H_{2}S\) at equilibrium are reported as this value. Similarly, the moles of \(CS_{2}\) at equilibrium are \(y\), which is equal to \(1.42 \times 10^{-3}\), as indicated by the moles of \(BaSO_4\). As a result, there are two equations: \(2x - 2y = 9.54 \times 10^{-3}\) and \(y = 1.42 \times 10^{-3}\). Solving these two equations will give us the values of \(x\) and \(y\).

Calculation of the Pressure at Equilibrium

With the total pressure of 1 atm, we need to distribute this based on the partial pressures of the molecules to ensure that the total pressure remains unchanged. The partial pressure of a molecule is proportional to the mole fraction of that molecule in the system. Therefore, the mole fraction of each molecule needs to be calculated, which will then allow us to find the partial pressures, a prerequisite to finding \(K_p\). The mole fraction is the moles of a molecule divided by the total moles in the system.

Calculation of the Equilibrium Constant \(K_p\)

Now that the partial pressures of all substances at equilibrium are determined, we can substitute these into the expression for \(K_p\), which is the ratio of the product of the pressures of the products (each raised to their stoichiometric coefficients) to the product of the pressures of the reactants (each raised to their stoichiometric coefficients). Be careful to assign the stoichiometric coefficients correctly. For this reaction, the expression for \(K_p\) is [P(CS2)][P(H2)]^4 / ([P(H2S)]^2 [P(CH4)]). Substituting the calculated partial pressures into this expression will provide us with the value of \(K_p\) at \(700^{\circ}C\).

Key concepts

Equilibrium Constant Kp

Understanding chemical equilibrium is key to grasping how reactions proceed under certain conditions. An important aspect of this is the equilibrium constant (Kp), which is a measure of the extent of a reaction at equilibrium at a given temperature, using the partial pressures of gases in the reaction. Kp is a dimensionless quantity that helps us predict the direction of the reaction: whether the reactants or the products are favored.

• Calculating Kp: Kp is calculated from the equilibrium concentrations of the reactants and products, each raised to the power of their coefficients in the balanced chemical equation.
• Expressing Kp: For the reaction aA + bB ⇌ cC + dD, Kp would be expressed as \( K_p = \frac{\left( P_C \right)^c \cdot \left( P_D \right)^d}{\left( P_A \right)^a \cdot \left( P_B \right)^b} \), where P represents the partial pressure of each component.
• Implications of Kp: A large Kp value indicates that the products are favored at equilibrium, while a small value means the reactants are favored. If Kp is close to 1, the system is at a relatively balanced state of equilibrium.

For an equation like our exercise, where 2 H₂S(g) + CH₄(g) ⇌ CS₂(g) + 4 H₂(g), we would use the partial pressures and stoichiometry of the equation to find Kp. This process involves using the given equilibrium conditions and the stoichiometry of the reaction to determine the partial pressures of each gas, and then applying those values to the formula for Kp.

Partial Pressure

In a mixture of gases, the partial pressure of a single type of gas is the pressure it would exert if it alone occupied the entire volume of the mixture at the same temperature. The concept of partial pressure is critical when dealing with gases at equilibrium because it reflects how much of each gas is present and how it contributes to the overall pressure of the system.

• Dalton's Law of Partial Pressures: This law states that the total pressure of a gas mixture is the sum of the partial pressures of each gas in the mixture.
• Calculating Partial Pressure: If you know the total pressure and the mole fraction of the gas in the mixture, you can calculate the partial pressure of that gas using \( P_{\text{gas}} = X_{\text{gas}} \cdot P_{\text{total}} \), where \(X_{\text{gas}}\) is the mole fraction and \(P_{\text{total}}\) is the total pressure.

In our example, the total pressure is given as 1 atm, and by calculating the mole fraction of each gas at equilibrium, we can then determine its partial pressure. This step is essential for computing the Kp of the reaction.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. Defined as the ratio of the number of moles of a component to the total number of moles of all components in the mixture, mole fractions are dimensionless numbers that offer a way to relate the amounts of different substances present in the system.

• Computing Mole Fractions: To calculate a mole fraction, divide the number of moles of the component of interest by the total number of moles in the mixture.
• Mole Fraction in Equilibrium Calculations: In terms of equilibrium calculations, mole fractions are used to determine the partial pressures, which in turn are used to calculate the equilibrium constant Kp.

When we have the total moles of all gases present at equilibrium, the mole fraction can be easily found, which will be crucial in finding the partial pressures needed for the equilibrium calculations.

Equilibrium Calculation

To solve equilibrium problems, we must use what's known as an equilibrium calculation. This involves a series of steps that typically start with setting up the balanced equation and the initial conditions before applying the known values to solve for the unknowns at equilibrium.

• Stoichiometry of the Reaction: A comprehensive understanding of the stoichiometry of the reaction is essential for determining the changes that occur as the system moves towards equilibrium.
• ICE Table: An ICE (Initial, Change, Equilibrium) table is often helpful for visualizing the shifts in concentrations or pressures of reactants and products as the reaction proceeds.
• Solving for Equilibrium: By analyzing the known and unknown variables, one can set up and solve mathematical equations that represent the relationships between the concentrations or partial pressures of the reactants and products.

As illustrated by the steps in our provided exercise solution, equilibrium calculations often require setting up a system of equations that reflect the conservation of mass and the balanced chemical equation. Once the equations are solved, one can then find the equilibrium concentrations or pressures and finally calculate the equilibrium constant Kp, which is a unique value characterizing the system at a specific temperature.