Question

The equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) is estab- lished at \(500 \mathrm{~K}\). An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for \(\mathrm{NO}, \mathrm{Cl}_{2},\) and \(\mathrm{NOCl}\), respectively. (a) Calculate \(K_{p}\) for this reaction at \(500.0 \mathrm{~K}\). (b) If the vessel has a volume of \(5.00 \mathrm{~L}\) calculate \(K_{c}\) at this temperature.

Short answer

(a) The equilibrium constant \(K_p\) for this reaction at 500.0 K is approximately 59.33. (b) The equilibrium constant \(K_c\) for this reaction at 500.0 K and a volume of 5.00 L is approximately 8.30.

Step-by-step solution

Write the equilibrium expression for Kp.

The equilibrium expression for Kp is given by: \( K_p = \frac{(P_{\mathrm{NOCl}})^2}{(P_{\mathrm{NO}})^2 * (P_{\mathrm{Cl_{2}}}) }\) Where P_NOCl, P_NO, and P_Cl2 are the partial pressures of NOCl, NO, and Cl2, respectively, at equilibrium.

Calculate Kp using the given partial pressures.

The given partial pressures for the gases at equilibrium are: P_NO = 0.095 atm, P_Cl2 = 0.171 atm, and P_NOCl = 0.28 atm. Substituting these values into the equilibrium expression, we can calculate Kp. \( K_p = \frac{(0.28)^2}{(0.095)^2 * (0.171)} \) Now solve for Kp: \( K_p ≈ 59.33 \)

Convert partial pressures to concentrations.

To find Kc, we need to convert the partial pressures to concentrations. We can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Rearranging this equation to find the concentration, we get: \[ \mathrm{C} = \frac{\mathrm{n}}{\mathrm{V}} = \frac{\mathrm{P}}{\mathrm{RT}} \] The given volume of the vessel is 5.00 L and the temperature is 500 K. We also need to use the gas constant R = 0.0821 L atm/mol K. Now, let's find the concentrations for all the gases: \( C_{\mathrm{NO}} = \frac{0.095}{(0.0821)(500)} ≈ 2.30 \times 10^{-3} M \) \( C_{\mathrm{Cl_2}} = \frac{0.171}{(0.0821)(500)} ≈ 4.16 \times 10^{-3} M \) \( C_{\mathrm{NOCl}} = \frac{0.28}{(0.0821)(500)} ≈ 6.79 \times 10^{-3} M \)

Write the equilibrium expression for Kc.

The equilibrium expression for Kc is given by: \( K_c = \frac{([\mathrm{NOCl}])^2}{([\mathrm{NO}])^2 * ([\mathrm{Cl_{2}}]) }\) Where [NOCl], [NO], and [Cl2] are the concentrations of NOCl, NO, and Cl2, respectively, at equilibrium.

Calculate Kc using the calculated concentrations.

Substitute the calculated concentrations into the Kc expression: \( K_c = \frac{(6.79 \times 10^{-3})^2}{(2.30 \times 10^{-3})^2 * (4.16 \times 10^{-3})} \) Now solve for Kc: \( K_c ≈ 8.30 \)

Report the final answers.

(a) The equilibrium constant Kp for this reaction at 500.0 K is approximately 59.33. (b) The equilibrium constant Kc for this reaction at 500.0 K and a volume of 5.00 L is approximately 8.30.

Key concepts

Kp calculation

The equilibrium constant, represented as \(K_p\), reflects the ratio of the partial pressures of the products to the reactants, each raised to the power of their respective stoichiometric coefficients in a balanced chemical equation.
For the equilibrium reaction \(2\, \text{NO}(g) + \text{Cl}_2(g) \rightleftharpoons 2 \text{NOCl}(g)\), the expression is:

• \(K_p = \frac{(P_{\text{NOCl}})^2}{(P_{\text{NO}})^2 \cdot (P_{\text{Cl}_2})}\)

The calculation of \(K_p\) involves substituting the given partial pressures into this expression. Here, \(P_{\text{NOCl}} = 0.28\, \text{atm}\), \(P_{\text{NO}} = 0.095\, \text{atm}\), and \(P_{\text{Cl}_2} = 0.171\, \text{atm}\).
Substitute these values to find:
\[K_p = \frac{(0.28)^2}{(0.095)^2 \cdot (0.171)} \approx 59.33\]
This value is a measure of the position of equilibrium at a given temperature and pressure.

Kc calculation

To compute the equilibrium constant in terms of concentration, \(K_c\), the partial pressures are first converted to molar concentrations using the Ideal Gas Law. This shift from pressures to concentrations is crucial for chemists working with reactions in solutions.
Apply the Ideal Gas Law, \(PV = nRT\), rearranged to express concentration \(C = \frac{P}{RT}\).

• Use \(R = 0.0821\, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}\) and \(T = 500\, \text{K}\).
• Convert each partial pressure to molarity:
  \(C_{\text{NO}} \approx \frac{0.095}{(0.0821)(500)} \approx 2.30 \times 10^{-3}\, \text{M}\)
  \(C_{\text{Cl}_2} \approx \frac{0.171}{(0.0821)(500)} \approx 4.16 \times 10^{-3}\, \text{M}\)
  \(C_{\text{NOCl}} \approx \frac{0.28}{(0.0821)(500)} \approx 6.79 \times 10^{-3}\, \text{M}\)

Then substitute these concentrations into the \(K_c\) expression:
\[K_c = \frac{(6.79 \times 10^{-3})^2}{(2.30 \times 10^{-3})^2 \cdot (4.16 \times 10^{-3})} \approx 8.30\]
\(K_c\) thus provides an alternative measure of equilibrium, ideal for reactions in liquid solutions.

Equilibrium expression

An equilibrium expression is a mathematical way of representing the concentrations or partial pressures of reactants and products in a chemical equilibrium.
It is unique to each reaction and derived from the balanced equation, considering stoichiometric coefficients.
For a general reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium expression for the equilibrium constant \(K\) is:

• \(K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\) or \(K = \frac{P_C^cP_D^d}{P_A^aP_B^b}\)

Where brackets \([]\) denote concentration and \(P\) represents partial pressures.
The expression helps determine the extent of the reaction under equilibrium conditions.

Partial pressures

Partial pressure is the pressure exerted by a single gas in a mixture of gases. It is a component of the total pressure within a closed system.
Each gas in a mixture contributes to the total pressure according to its proportion in the mixture.
Dalton's Law of Partial Pressures helps us understand this concept, stating that the total pressure of a gas mixture is equal to the sum of the partial pressures of the constituent gases:

• \(P_{\text{total}} = P_1 + P_2 + P_3 + \ldots\)

In chemical equilibria, partial pressures are essential as they are used directly in expressions to calculate \(K_p\).
Knowing the partial pressures of reactants and products helps establish the position of balance between forward and reverse reactions.

Ideal gas law

The Ideal Gas Law is a vital equation in chemistry that links pressure, volume, temperature, and moles of gas. It is articulated as \(PV = nRT\), where:

• \(P\) is the pressure
• \(V\) is the volume
• \(n\) is the amount of substance in moles
• \(R\) is the ideal gas constant (0.0821 L atm/mol K)
• \(T\) is the temperature in Kelvin

This law is pivotal for deriving concentrations from partial pressures, particularly crucial for transitioning from \(K_p\) to \(K_c\) calculations.
It assumes gases behave ideally, which means real-life gases align closely to calculations only under certain conditions (low pressure and high temperature).
As a fundamental concept in chemistry, it aids in predicting how gases behave in various physical processes.