Problem 20 - Chemistry: The Central Science

Question

Consider the following equilibrium: \(2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) \quad K_{c}=1.08 \times 10^{7}\) at \(700{ }^{\circ} \mathrm{C}\) (a) Calculate \(K_{p}\). (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly \(\mathrm{H}_{2} \mathrm{~S}\) ?

Step-by-Step Solution

Answer

(a) The Kp of the given reaction is \(1.35 \times 10^{5}\). (b) The equilibrium mixture contains mostly H2S and very little H2 and S2.

Step 1: Relationship between Kc and Kp

Using the relationship between Equilibrium Constants Kc and Kp: \[K_p = K_c(RT)^{(\Delta n)}\] where R is the ideal gas constant (0.08206 L atm/mol K), T is the temperature in Kelvin, and Δn is the change in the number of moles of gaseous species in the balanced chemical equation.

Step 2: Convert temperature from Celsius to Kelvin

To convert the given temperature from Celsius to Kelvin, add 273.15 to the Celsius temperature: \[T(K) = T(^{\circ}C) + 273.15\] \[T(K) = 700 + 273.15 = 973.15\, K\]

Step 3: Calculate the change in the number of moles of gaseous species, Δn

Looking at the balanced chemical equation: \[2\,H_2(g) + S_2(g) \rightleftharpoons 2\,H_2S(g)\] The change in the number of moles of gaseous species is Δn = (moles of products) – (moles of reactants): \[\Delta n = 2 - (2 + 1) = -1\]

Step 4: Calculate Kp using Kc, R, T, and Δn

Now, we can calculate Kp using the given Kc, the ideal gas constant R, the temperature T in Kelvin, and the calculated Δn: \[K_p = K_c(RT)^{(\Delta n)}\] \[K_p = (1.08 \times 10^7)(0.08206 \times 973.15)^{-1}\] \[K_p = (1.08 \times 10^7)(79.94)^{-1}\] \[K_p = \frac{1.08 \times 10^7}{79.94}\]

Step 5: Simplify Kp

Solve for Kp: \[K_p = 1.35 \times 10^5\] Now that we have Kp, let's analyze the composition of the equilibrium mixture.

Step 6: Analyze the composition of the equilibrium mixture

Equilibrium constant Kc is given as \(1.08 \times 10^7\). This large value indicates that the reaction proceeds mostly to the right, producing more H2S. As a result, the equilibrium mixture contains mostly H2S and very little H2 and S2. In conclusion, (a) The Kp of the given reaction is \(1.35 \times 10^{5}\). (b) The equilibrium mixture contains mostly H2S and very little H2 and S2.

Key Concepts

Equilibrium Constants

Understanding chemical equilibrium is crucial for interpreting how chemical reactions behave under static conditions. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products over time. This state is numerically defined by an equilibrium constant, which is a value representing the ratio of concentration terms for products to reactants, each raised to the power of their coefficients in the balanced chemical equation.

For instance, in the exercise, the reaction 2 H₂(g) + S₂(g) ⇌ 2 H₂S(g) has an equilibrium constant, K_c, equal to 1.08 × 10⁷. Such a high K_c value suggests that, when equilibrium is reached, the concentration of the products, in this case, H₂S, is significantly greater than the concentrations of the reactants, implying the reaction heavily favors product formation.

Kc and Kp Relationship

Chemical equilibria for gaseous reactions can be expressed using either K_c or K_p. K_c is based on the concentration (molarity) of substances, while K_p relates to the partial pressures of gases. The relationship between K_c and K_p for any gaseous equilibrium is given by the equation:
\[K_p = K_c(RT)^{(\triangle n)}\]
R is the ideal gas constant and T is the temperature in Kelvin. Δn is the change in moles of gas particles when proceeding from reactants to products.

In the provided exercise, by completing the conversion from °C to K and calculating Δn, we use these values to transition from K_c to K_p. This process not only solidifies understanding of the connection between these constants but also highlights the direct influence temperature and molar changes have on the position of equilibrium.

Le Chatelier's Principle

Le Chatelier's principle provides a qualitative understanding of how a system at equilibrium responds to external changes, such as alterations in concentration, pressure, or temperature. According to this principle, if an external stress is imposed on a system at equilibrium, the system adjusts itself in such a way as to counteract the imposed change and re-establish a new equilibrium state.

For example, if the concentration of a reactant is increased, the system shifts to consume that excess reactant by making more products. Conversely, reducing the pressure over a gaseous system by increasing its volume results in shifting the equilibrium toward the side with more moles of gas to increase pressure.

The exercise, does not directly involve Le Chatelier's principle, concentrating instead on the calculation aspect. However, understanding this principle is instrumental for predicting how the equilibrium will shift in response to various stresses, which in turn affects both K_c and K_p values and the composition of an equilibrium mixture.