Question

Consider the following equilibrium, for which \(K_{p}=0.0752\) at \(480^{\circ} \mathrm{C}\) : \(2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g)\) (a) What is the value of \(K_{p}\) for the reaction \(4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) ?\) (b) What is the value of \(K_{p}\) for the reaction \(\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?\) (c) What is the value of \(K_{c}\) for the reaction in part (b)?

Short answer

(a) The value of Kp for the reaction \(4\mathrm{HCl}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2\mathrm{Cl}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(g)\) is 13.2979. (b) The value of Kp for the reaction \(\mathrm{Cl}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(g) \rightleftharpoons 2\mathrm{HCl}(g) + \frac{1}{2}\mathrm{O}_{2}(g)\) is 3.6458. (c) The value of Kc for the reaction \(\mathrm{Cl}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(g) \rightleftharpoons 2\mathrm{HCl}(g) + \frac{1}{2}\mathrm{O}_{2}(g)\) is 0.7445.

Step-by-step solution

1. Determining Kp of the reversed reaction:

To obtain the reversed reaction in (a), change the direction of the given reaction as: \( 4\mathrm{HCl}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2\mathrm{Cl}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(g)\). The relationship between the equilibrium constants of the original and reversed reaction is: \( K'_{p} = \frac{1}{K_{p}}\). Since \(K_{p} = 0.0752\), we can easily calculate the value of \(K'_{p}\) as follows: \[ K'_{p} = \frac{1}{0.0752} \]

2. Calculating Kp for the reversed reaction:

Now, let's compute the value of Kp for the reversed reaction: \[ K'_{p} = \frac{1}{0.0752} = 13.2979 \]

3. Determining Kp of the halved reaction:

From the reversed reaction, let's divide the stoichiometric coefficients by 2 to get the desired reaction in (b): \(\mathrm{Cl}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(g) \rightleftharpoons 2\mathrm{HCl}(g) + \frac{1}{2}\mathrm{O}_{2}(g)\). When a reaction is multiplied or divided by a factor, the relationship between the equilibrium constants is as follows: \( K''_{p} = (K'_{p})^{\frac{1}{n}} \) , where n is the factor we have divided the reaction. Here, \(n = 2\), and we can calculate the new Kp value as follows: \[ K''_{p} = (13.2979)^{\frac{1}{2}} \]

4. Calculating Kp for the halved reaction:

Now, let's compute the value of Kp for the halved reaction: \[ K''_{p} = (13.2979)^{\frac{1}{2}} = 3.6458 \]

5. Determining Kc from Kp:

To find the Kc value from the Kp value, we can use the relationship: \( K_{c} = K_{p} \times (RT)^{-\Delta n}\), where R is the ideal gas constant (0.0821 L⋅atm/mol⋅K), T is the temperature in Kelvin, and \(\Delta n\) is the difference in the moles of gaseous products and reactants. For the reaction in part (b), the temperature is given as 480°C, which is equivalent to 753.15 K, and the mole difference is (2 - 1.5) = 0.5.

6. Calculate Kc for the halved reaction:

Now we can calculate Kc for the halved reaction using the previously mentioned relationship: \[ K_{c} = K''_{p} \times (RT)^{-\Delta n} = 3.6458 \times (0.0821 \times 753.15)^{-0.5} \] Let's compute the value of Kc: \[ K_{c} = 3.6458 \times (0.0821 \times 753.15)^{-0.5} = 0.7445 \]

Summary:

(a) The value of Kp for the reaction \(4\mathrm{HCl}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2\mathrm{Cl}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(g)\) is 13.2979. (b) The value of Kp for the reaction \(\mathrm{Cl}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(g) \rightleftharpoons 2\mathrm{HCl}(g) + \frac{1}{2}\mathrm{O}_{2}(g)\) is 3.6458. (c) The value of Kc for the reaction \(\mathrm{Cl}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(g) \rightleftharpoons 2\mathrm{HCl}(g) + \frac{1}{2}\mathrm{O}_{2}(g)\) is 0.7445.

Key concepts

Equilibrium Constant (Kp)

Understanding the equilibrium constant (Kp) is essential in the study of chemical reactions and their behavior at equilibrium. Kp represents the ratio of partial pressures of the products to reactants in a gaseous system at equilibrium, raised to the power of their respective coefficients in the balanced equation. It is a measure of the extent to which a reaction proceeds before reaching a state of balance.

For the reaction given in the exercise, the Kp value indicates the position of equilibrium at a specific temperature. When the reaction is reversed or altered, as in parts (a) and (b) of the exercise, we apply mathematical relationships to derive the new equilibrium constants. Importantly, for a reverse reaction, Kp is simply the reciprocal of the original value, and for a reaction multiplied or divided by a number, we use the nth root or power of the initial Kp, respectively.

Reaction Quotient

While the equilibrium constant (Kp) is a fixed value at a specific temperature, the reaction quotient (Q) describes the ratio of concentrations or pressures of the reactants and products at any point during the reaction, before equilibrium is reached. Q is calculated using the same formula as Kp, but with the current concentrations or partial pressures.

Comparing Q to Kp helps predict the direction in which the reaction will proceed to reach equilibrium. If Q is less than Kp, the forward reaction is favored, and more products will form. If Q exceeds Kp, the reaction will move in the reverse direction. This approach is crucial for predicting changes in reaction conditions and is intimately connected to Le Chatelier's principle, discussed in upcoming sections.

Le Chatelier's Principle

Le Chatelier's principle is a cornerstone concept in chemistry that helps predict how a system at equilibrium responds to changes in concentration, pressure, or temperature.

According to this principle, when a stress is applied to a system at equilibrium, the system will adjust to minimize the stress and restore a state of balance. For instance, increasing the concentration of reactants will drive the reaction forward to produce more products. Conversely, increasing the product concentrations will shift the equilibrium towards reactants. Likewise, changes in pressure and temperature will also result in predictable shifts of the equilibrium position, emphasizing the dynamic nature of chemical equilibrium.

Equilibrium Expressions

Equilibrium expressions are the mathematical representations used to denote the equilibrium constant for a reaction. For reactions involving gases, the equilibrium expression is written in terms of partial pressures (Kp), and for reactions involving solutes in solution, concentrations (Kc) are used.

In the given textbook exercise, the equilibrium expression for Kp is calculated based on the partial pressures of the gaseous species. The steps reveal the importance of following the stoichiometry of the balanced equation, with exponents corresponding to the stoichiometric coefficients. A reversed or altered reaction requires adjustments in the equilibrium expression to reflect the correct stoichiometry and maintain the thermodynamic relationship between the reactants and products.