Problem 75 - Chemistry and Chemical Reactivity

Question

Suppose a tank initially contains \(\mathrm{H}_{2} \mathrm{S}\) at a pressure of 10.00 atm and a temperature of 800 K. When the reaction has come to equilibrium, the partial pressure of \(\mathbf{S}_{2}\) vapor is 0.020 atm. Calculate \(K_{\mathrm{p}}\) $$ 2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{S}_{2}(\mathrm{g}) $$

Step-by-Step Solution

Answer

The equilibrium constant \(K_{\mathrm{p}}\) is approximately \(3.2 \times 10^{-7}\).

Step 1: Write the Equation for Kp

The equilibrium constant in terms of partial pressures, \(K_{\mathrm{p}}\), for the reaction \(2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{S}_{2}(\mathrm{g})\) is given by: \[ K_{\mathrm{p}} = \frac{(P_{\mathrm{H}_{2}})^2 \cdot P_{\mathrm{S}_{2}}}{(P_{\mathrm{H}_{2}\mathrm{S}})^2} \] Where \(P_{\mathrm{H}_{2}}\), \(P_{\mathrm{S}_{2}}\), and \(P_{\mathrm{H}_{2}\mathrm{S}}\) are the equilibrium partial pressures of \(\mathrm{H}_{2}\), \(\mathrm{S}_{2}\), and \(\mathrm{H}_{2}\mathrm{S}\) respectively.

Step 2: Set Up Initial and Change Table for Pressures

Initially, the pressure of \( \mathrm{H}_{2}\mathrm{S}\) is 10.00 atm and \(\mathrm{H}_{2}\) and \( \mathrm{S}_{2} \) are absent. As \( x \) atm of \( \mathrm{S}_{2} \) forms, the pressure of \( \mathrm{H}_{2} \) will be \( 2x \) atm, and the pressure of \( \mathrm{H}_{2}\mathrm{S} \) will decrease by \( 2x \) atm. The equilibrium pressures are: - \( P_{\mathrm{S}_{2}} = 0.020 \; \text{atm} \) (given) - \( P_{\mathrm{H}_{2}} = 2x \) atm - \( P_{\mathrm{H}_{2}\mathrm{S}} = 10.00 - 2x \) atm Where \( x = 0.020 \; \text{atm} \).

Step 3: Calculate Equilibrium Pressures of Components

Substitute \( x = 0.020 \) atm into pressures: - \( P_{\mathrm{S}_{2}} = 0.020 \; \text{atm} \) - \( P_{\mathrm{H}_{2}} = 2 \times 0.020 = 0.040 \; \text{atm} \) - \( P_{\mathrm{H}_{2}\mathrm{S}} = 10.00 - 2(0.020) = 9.96 \; \text{atm} \)

Step 4: Substitute Pressures into Kp Expression

Insert the equilibrium pressures into the expression for \( K_{\mathrm{p}} \): \[ K_{\mathrm{p}} = \frac{(0.040)^2 \times 0.020}{(9.96)^2} \]

Step 5: Solve for Kp

Calculate the value of \( K_{\mathrm{p}} \): \[ K_{\mathrm{p}} = \frac{0.0016 \times 0.020}{99.20016} = \frac{0.000032}{99.20016} \approx 3.2 \times 10^{-7} \]

Key Concepts

Partial Pressure

Partial pressure is a concept often used in chemical equations involving gases. It refers to the pressure that a single type of gas exerts in a mixture of gases. The partial pressure of a gas is a measure of its concentration in a mixture. It is important because it allows you to calculate the behavior of gases in chemical reactions using substances' noting how they directly relate to overall reaction dynamics.
For each gas in a mixture, its partial pressure is proportional to its mole fraction in the mixture. This can be calculated by the equation:

• Partial Pressure: \( P_i = X_i \times P_{total} \)

where \( P_i \) is the partial pressure of gas \( i \), \( X_i \) is its mole fraction, and \( P_{total} \) is the total pressure of the gas mixture.
In our exercise, initially, you only have \(P_\mathrm{H_{2}S}\), which is 10.00 atm. As the reaction progresses to equilibrium, the formation of \(S_2\) results in changes that help us establish other partial pressures needed, such as \(P_\mathrm{H_2}\). Understanding these concepts makes it easier to proceed with calculating equilibrium constants.

Chemical Equilibrium

Chemical equilibrium refers to the state in a chemical reaction at which the concentrations of reactants and products remain constant over time. At this point, the forward and reverse reactions occur at equal rates. In other words, there is no net change in the amounts of substances involved. This does not mean that reactions have stopped; rather, they are proceeding equally fast in both directions.
For our reaction \( 2\text{H}_2\text{S}\rightleftharpoons 2\text{H}_2 + \text{S}_2 \), equilibrium is reached when the formation rates of \( \text{H}_2 \) and \( \text{S}_2 \) match their respective depletion rates. This dynamic balance implies the partial pressures or concentrations of all reactants and products remain consistent.
The equilibrium constant \( K_p \), used in this exercise, expresses the ratio of products' partial pressures over reactants', each raised to the power of their stoichiometric coefficients. The expression looks like this:

• Equilibrium Constant: \( K_p = \frac{(P_\mathrm{H_{2}})^2 \cdot P_\mathrm{S_{2}}}{(P_\mathrm{H_{2}S})^2} \)

Being able to calculate and understand \( K_p \) from given partial pressures showcases the chemical system in equilibrium, allowing predictions about the direction of the reaction under different conditions.

Reaction Kinetics

Reaction kinetics is the study of the speed or rate at which chemical reactions occur. This includes understanding how different conditions affect these rates, such as concentrations, temperatures, and catalysts.
In our given reaction, we observe how the initial amount of \( \text{H}_2\text{S} \) translates to shifts in pressure as the equilibrium state forms. Reaction kinetics provides insight into how quickly equilibrium is reached and how factors like temperature (in this case, 800 K) alter the speed of reaction progression.
A key characteristic in kinetics is the idea that reactions speed up with increasing temperature, which is consistent with the collision theory. This theory posits that higher temperatures increase the energy and frequency of collisions between reactant molecules, boosting the likelihood of successful reactions.
For equilibrium and kinetics, it's pivotal to grasp how the reaction mechanism, which describes step-by-step transformation of reactants to products, affects the overall reaction speed towards equilibrium. Understanding these kinetics aspects ensures predictable control over industrial and laboratory reactions, affecting efficiencies for desired products.