Question

Consider the following equilibrium: \(2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{S}(g) \quad K_{c}=1.08 \times 10^{7} \mathrm{at} 700^{\circ} \mathrm{C}\) (a) Calculate \(K_{p}\) . (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly \(\mathrm{H}_{2} \mathrm{S} ?(\mathbf{c})\) Calculate the value of \(K_{c}\) if you rewrote the equation \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{S}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{S}(g)\)

Short answer

In summary, we have calculated the equilibrium constant \(K_p\) for the given reaction to be 0.042. The equilibrium mixture consists predominantly of hydrogen sulfide (H₂S) rather than hydrogen (H₂) and sulfur (S₂). Finally, the value of Kc for the rewritten equilibrium equation is approximately 3,286.

Step-by-step solution

Understand the relationship between Kc and Kp

The relationship between Kc and Kp is given by the equation: \[K_p = K_c(RT)^{\Delta n}\] where R is the ideal gas constant, T is the temperature in Kelvin, and Δn is the difference in the number of moles of gas between the products and the reactants. For the given equilibrium: \[2 \mathrm{H}_{2}(g) + \mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2}\mathrm{S}(g)\] Δn = (2 moles of H2S) - (2 moles of H2 + 1 mole of S2) = -1. Now we can proceed to calculate Kp.

Convert temperature to Kelvin and find Kp

First, convert the temperature from Celsius to Kelvin: T = 700 + 273.15 = 973.15 K Next, use the relationship between Kc and Kp to find Kp: \[K_p = K_c(RT)^{\Delta n} = (1.08 \times 10^7)(8.314 \times 973.15)^{-1} = 0.042\] So, \(K_p = 0.042\). For part (b), we will analyze the equilibrium mixture.

Determine the predominant components of the equilibrium mixture

Since \(K_c\) is greater than 1 (\(K_c = 1.08 \times 10^7\)), this means that the equilibrium favors the product, hydrogen sulfide (H2S). Therefore, the equilibrium mixture contains mostly H2S rather than H2 and S2. For part (c), we will calculate the value of Kc for the modified equilibrium equation.

Understand the relationship between the original and modified Kc values

The original equilibrium equation can be rewritten as twice the modified equation: \(\begin{array}{cccc} 2\mathrm{H}_{2}(g)&+&\mathrm{S}_{2}(g)&\rightleftharpoons2\mathrm{H}_{2}\mathrm{S}(g)\\ 2(\mathrm{H}_{2}(g)&+\frac{1}{2}\mathrm{S}_{2}(g)&\rightleftharpoons\mathrm{H}_{2}\mathrm{S}(g))& \end{array}\) We can denote the original Kc value as \(K_{c1}\) and the modified Kc value as \(K_{c2}\). Since the second reaction is equal to half of the first reaction, their equilibrium constants are related by the expression: \[K_{c1} = K_{c2}^2\] Now we can calculate the modified Kc value.

Calculate the modified Kc value

Use the relationship found in Step 4 to solve for the modified Kc value: \[K_{c2} = \sqrt{K_{c1}} = \sqrt{1.08 \times 10^7} = 3,286\] So, the value of Kc for the modified equilibrium equation is approximately 3,286.

Key concepts

Kp calculation

When dealing with gas-phase reactions at equilibrium, it is essential to distinguish between the equilibrium constants for concentrations (\(K_c\)) and pressures (\(K_p\)). Calculating \(K_p\) involves using the relationship between \(K_c\) and \(K_p\), expressed as:\[K_p = K_c(RT)^{\Delta n}\]Where:

• \(R\) is the ideal gas constant, typically 8.314 J/mol·K.
• \(T\) is the absolute temperature in Kelvin.
• \(\Delta n\) is the change in moles of gas from reactants to products.

For the example reaction:\[2 \mathrm{H}_2(g) + \mathrm{S}_2(g) \rightleftharpoons 2 \mathrm{H}_2\mathrm{S}(g)\]The change in moles, \(\Delta n\), is calculated as:\[\Delta n = 2 - (2 + 1) = -1\]This equation shows us the decrease in the number of moles from reactants to products. With the given \(K_c = 1.08 \times 10^7\) and converted temperature of 973.15 K, plug in the values:\[K_p = (1.08 \times 10^7) (8.314 \times 973.15)^{-1}\]The calculation reveals that \(K_p\) equals approximately 0.042, indicating the extent to which the reactants are converted to products under pressure conditions.

Kc and Kp relationship

Understanding the relationship between \(K_c\) and \(K_p\) is crucial for analyzing equilibrium in reactions where gases are involved. While \(K_c\) pertains to concentrations, \(K_p\) relates to partial pressures. These constants are interconnected through the formula:\[K_p = K_c(RT)^{\Delta n}\]This formula accounts for the generalized Ideal Gas Law where pressure and concentration are tied through temperature and volume.

• The constant \(R\) serves as the bridge between different units, ensuring that temperature \(T\) appropriately adjusts the balance of moles shifting from one side of the reaction to the other.
• The term \(\Delta n\) lets us account for how many moles of gas molecules are gained or lost as the reaction reaches equilibrium.
• Temperature, detailed in Kelvin, inflates or deflates these values proportionate to how far shifted the reaction is towards products or reactants.

Consequently, the conversion between \(K_c\) and \(K_p\) informs us about the affinity of the gaseous reaction participants under different circumstances of volume and pressure.

Equilibrium mixture analysis

Examining the equilibrium mixture involves understanding which side of the reaction is favored. In our exercise, the equilibrium constant \(K_c = 1.08 \times 10^7\) is substantially larger than 1, implying that the reaction favors the formation of products. This means hydrogen sulfide (\(\mathrm{H}_2\mathrm{S}\)) is predominantly present over reactants hydrogen (\(\mathrm{H}_2\)) and sulfur (\(\mathrm{S}_2\)). The magnitude of \(K_c\) directly indicates product dominance:

• A \(K_c\) greater than 1 shows product-favorable reactions.
• A \(K_c\) less than 1 would suggest reactant-favorable conditions.
• In our case, a very high \(K_c\) indicates almost complete conversion to products.

Equilibrium analysis often provides insights into which components could be better exploited or conserved in industrial processes, particularly when dealing with high-value molecules formed predominantly in the products.

Ideal gas constant

The ideal gas constant \(R\), integral in the formula relating \(K_c\) and \(K_p\), plays a pivotal role in equilibrium calculations. It is a constant used in the Ideal Gas Law and helps translate conditions of concentration to pressure and vice versa. The commonly accepted value for \(R\) is 8.314 J/mol·K.

• This constant ensures calculations remain consistent when scaling up laboratory findings to industrial or environmental conditions.
• The presence of \(R\) in the equations allows temperature to be a unifying factor in converting concentration into pressure through \(K_c\) and \(K_p\).
• By maintaining unit roles, \(R\) helps illustrate how different parameters interchange under thermodynamic scenarios.

Understanding \(R\) enables a seamless transition between theoretical calculations and practical applications, correctly predicting how gaseous reactions will behave under various thermal and pressure conditions.