Question

The equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) is established at \(500 \mathrm{~K}\). An equilibrium mixture of the three gases has partial pressures of \(0.095 \mathrm{~atm}, 0.171 \mathrm{~atm}\), and \(0.28 \mathrm{~atm}\) 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

\(K_p \approx 8.31\) at 500 K, and \(K_c \approx 83.25\) at 500 K.

Step-by-step solution

Write the balanced equation

The balanced equation is already given as: \[2 \mathrm{NO}(g) + \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\]

Calculate K_p using given partial pressures

To calculate \(K_p\), we can use the following formula: \[K_p = \frac{(\text{partial pressure of products})^{\text{stoichiometric coefficients}}}{(\text{partial pressure of reactants})^{\text{stoichiometric coefficients}}}\] For this reaction, we have: \[\begin{aligned} K_p &= \frac{(\text{pressure of NOCl})^2}{(\text{pressure of NO})^2 \times \text{pressure of Cl}_2} \\ &= \frac{(0.28 \text{ atm})^2}{(0.095 \text{ atm})^2 \times (0.171 \text{ atm})} \end{aligned}\]

Solve for K_p

Now we can solve for \(K_p\): \(K_p = \frac{(0.28)^2}{(0.095)^2 \times 0.171} \approx 8.31\) So, \(K_p \approx 8.31\) at 500 K. #b) Calculate K_c given the volume#

Convert partial pressures to concentrations using the Ideal Gas Law

We are given the volume of the vessel as 5.00 L. Using the Ideal Gas Law, we can convert partial pressures to concentrations: \[PV = nRT\] Rearrange to find the concentration: \[\text{Concentration} = \frac{n}{V} = \frac{P}{RT}\] At 500 K, use this formula to find the concentration of each species: \[\begin{aligned}\text{Concentration of NO} &= \frac{0.095\;\text{atm}}{\text{(0.0821 L.atm/mol.K)(500 K)}} \approx 0.00232\;\text{mol/L}\\ \text{Concentration of Cl}_2 &= \frac{0.171\;\text{atm}}{\text{(0.0821 L.atm/mol.K)(500 K)}} \approx 0.00416\;\text{mol/L}\\ \text{Concentration of NOCl} &= \frac{0.28\;\text{atm}}{\text{(0.0821 L.atm/mol.K)(500 K)}} \approx 0.00680\;\text{mol/L}\end{aligned}\]

Calculate K_c using the calculated concentrations

Now, use the same formula as in Step 2, replacing partial pressures with concentrations to find \(K_c\): \[\begin{aligned} K_c &= \frac{(\text{concentration of NOCl})^2}{(\text{concentration of NO})^2 \times \text{concentration of Cl}_2} \\ &= \frac{(0.00680)^2}{(0.00232)^2 \times 0.00416} \end{aligned}\]

Solve for K_c

Now we can solve for \(K_c\): \(K_c = \frac{(0.00680)^2}{(0.00232)^2 \times 0.00416} \approx 83.25 \) So, \(K_c \approx 83.25\) at 500 K.

Key concepts

Understanding Equilibrium Constant (Kp)

Equilibrium in chemistry is a dynamic state where the rate of the forward reaction equals the rate of the reverse reaction, meaning the concentrations of reactants and products remain constant over time. A vital tool to quantify this state is the equilibrium constant, expressed in terms of partial pressures as the equilibrium constant (Kp). To calculate Kp, the partial pressures of the products, raised to their coefficients in the balanced equation, are divided by the partial pressures of the reactants, also raised to their respective coefficients. This ratio gives a value that is constant for a given reaction at a constant temperature, signifying the ratio of product pressures to reactant pressures at equilibrium.

For example, the reaction given in the exercise, after substituting the given partial pressures into the formula, yields an equilibrium constant, Kp, which provides a numerical value reflecting the extent of the reaction. A higher Kp value indicates a greater extent of product formation at equilibrium.

Determining Partial Pressures in Equilibrium

In an equilibrium mixture of gases, the partial pressure of each gas is a crucial concept. It refers to the pressure that the gas would exert if it alone occupied the entire volume of the mixture at the same temperature. The partial pressures at equilibrium are used to determine the equilibrium constant (Kp). These pressures are 'partial' because they represent a portion of the total pressure exerted by the entire gas mixture.

For instance, in the given reaction, the respective partial pressures of NO, Cl₂, and NOCl are fundamental in calculating Kp. It's important to note that these pressures must be measured when the system is at equilibrium to accurately reflect the proportions of reactants and products that the reaction naturally establishes.

Relation Between Kp and Kc

While Kp is used for equilibria involving gases, expressed in terms of partial pressures, Kc is the equilibrium constant for concentrations in molarity. There is a direct relationship between these two constants for a given reaction at a constant temperature. This relationship is described by the formula: \[Kp = Kc(RT)^{\triangle n}\] where R is the ideal gas constant, T is the temperature in Kelvin, and Δn is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants). This equation bridges the gap between the two constants, permitting conversions based on whether pressure or concentration is the chosen condition for describing the system's equilibrium status.

In the exercise, after finding Kp, Kc can be calculated by first converting partial pressures to concentrations using the ideal gas law and then applying the equilibrium constant formula regarding concentrations.

Ideal Gas Law in Chemical Equilibrium

The ideal gas law, represented by the equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature, is pivotal in connecting partial pressures and concentrations at equilibrium. To find the equilibrium constant in terms of concentration (Kc), concentrations of the gases involved must be known. These can be derived from their partial pressures while considering the volume and temperature of the system, as shown in the given exercise.

The gas law allows for the translation of partial pressures into mole per liter units (M), which are needed for the calculation of Kc. By dividing the partial pressure by RT, one obtains the concentration, which can then be plugged into the equilibrium expression with concentrations. This fundamental law provides a pathway to quantitatively describe the gaseous equilibria in a coherent manner that accounts for the ideal behavior of gases.