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 \(K_{p \cdot}\) (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly \(\mathrm{H}_{2} \mathrm{~S} ?(\mathbf{c})\) Calculatethevalue 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

(a) The value of \(K_p\) at \(700^{\circ} \mathrm{C}\) is \(2.68\times10^8\). (b) The equilibrium mixture mostly contains H2S. (c) The new value of \(K_c\) for the rewritten equation is \(3.286 \times 10^3\).

Step-by-step solution

Convert Kc to Kp

To convert the equilibrium constant Kc to Kp, we use the following equation: \(K_p = K_c(RT)^{\Delta n}\), where R is the ideal gas constant (0.0821 atm L/mol K), T is the temperature in kelvin, and Δn is the change in moles of gas (products - reactants). First, we need to convert the temperature from Celsius to Kelvin: \(T = 700 + 273.15 = 973.15K\) Next, we calculate Δn: \(\Delta n = (2 - (2 + 1)) = -1\) Now, we can calculate Kp: \(K_p = K_c (RT)^{\Delta n} = (1.08 \times 10^7)((0.0821)(973.15))^{-1}\)

Calculate Kp

By plugging the values into the Kp formula, we get: \(K_p = (1.08 \times 10^7)(0.0799)^{-1} \Rightarrow K_p = 2.68 \times 10^8\) So, the value of Kp at 700°C is \(2.68\times10^8\).

Determine the composition of the equilibrium mixture

To determine the composition of the equilibrium mixture, we can compare the Kc value with the stoichiometry of the reaction. A Kc value significantly larger than 1 indicates that the reaction favors the products, while a Kc value significantly smaller than 1 indicates that the reaction favors the reactants. Since the given Kc value is very large at \(1.08 \times 10^7\), we can conclude that the equilibrium mixture will mostly contain H2S.

Calculate the new Kc for the rewritten equation

When the reaction is rewritten as \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{~S}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)\), the stoichiometric coefficients are halved relative to the original reaction. To find the new Kc value, we must square root the original Kc. New Kc = \(\sqrt{1.08 \times 10^7} \Rightarrow\) New Kc = \(3.286 \times 10^3\) The new value of Kc for the rewritten equation is \(3.286 \times 10^3\).

Key concepts

Equilibrium Constant (Kc)

Chemical reactions often reach a point where the rate of the forward reaction equals the rate of the backward reaction. At this point, known as equilibrium, the concentrations of the reactants and products remain constant.
This balance is quantified by the equilibrium constant, represented as \(K_c\).
The equilibrium constant for a reaction is derived from the concentrations of the reactants and products at equilibrium. For instance, in the reaction \(2 \mathrm{H}_2(g) + \mathrm{S}_2(g) \rightleftharpoons 2 \mathrm{H}_2 \mathrm{S}(g)\), the expression for \(K_c\) is:
\[K_c = \frac{[\mathrm{H}_2 \mathrm{S}]^2}{[\mathrm{H}_2]^2[\mathrm{S}_2]}\]

If \(K_c\) is much greater than 1, the position of equilibrium lies towards the products, meaning the reaction favors the formation of products.
Conversely, if \(K_c\) is much less than 1, the reaction favors the reactants. In the provided exercise, \(K_c = 1.08 \times 10^7\) at \(700^{\circ}\)C is very large, indicating a significant amount of \(\mathrm{H}_2 \mathrm{S}\) compared to the reactants.

Equilibrium Constant (Kp)

When dealing with reactions involving gases, the equilibrium constant can also be expressed in terms of partial pressures, represented by \(K_p\).
Like \(K_c\), \(K_p\) provides essential insights into the position of equilibrium.
To convert \(K_c\) to \(K_p\), we use the formula: \[K_p = K_c(RT)^{\Delta n}\]
where:

• \(R\) is the ideal gas constant, approximately 0.0821 atm L/mol K
• \(T\) is the temperature in Kelvin
• \(\Delta n\) is the change in moles of gas (products minus reactants)

In the provided exercise, we calculated \(\Delta n\) as \(-1\) since the number of moles on the product side is less than that on the reactant side.
By substituting the values, we obtained \(K_p = 2.68 \times 10^8\), indicating that at 700°C, the equilibrium mixture highly favors the product formation as well.

Stoichiometry

Stoichiometry involves the quantitative relationships between the amounts of reactants and products in a chemical reaction.
It is essential for understanding how changes in reaction conditions affect equilibrium constants. In the rewritten reaction \(\mathrm{H}_2(g) + \frac{1}{2} \mathrm{S}_2(g) \rightleftharpoons \mathrm{H}_2 \mathrm{S}(g)\), the stoichiometry changes because the coefficients are altered.
When the stoichiometric coefficients are modified, it affects the equilibrium constant calculation.
In this exercise, changing the coefficients to half their original value requires taking the square root of the initial \(K_c\), resulting in a new \(K_c\) of \(3.286 \times 10^3\).
This adjustment helps us understand how even minor stoichiometric changes can impact the equilibrium constants and, consequently, the reaction dynamics.