Question

Consider the equilibrium: $$ 2 \mathrm{NOBr}(g) \rightleftarrows 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ If nitrosyl bromide (NOBr) is 34 percent dissociated at \(25^{\circ} \mathrm{C}\) and the total pressure is 0.25 atm, calculate \(K_{P}\) and \(K_{\mathrm{c}}\) for the dissociation at this temperature.

Short answer

\(K_P = 0.0298\), \(K_c = 0.001227\).

Step-by-step solution

Define Initial Conditions and Changes

Consider the initial moles of NOBr to be 1 mole. Since it is 34% dissociated, the change in moles is 0.34 mol. Therefore, at equilibrium, we have the following: - For NOBr: Initial = 1 mol, Change = -0.34 mol, Equilibrium = 0.66 mol - For NO: Initial = 0 mol, Change = +0.34 mol × 2 = +0.68 mol, Equilibrium = 0.68 mol - For Br₂: Initial = 0 mol, Change = +0.17 mol, Equilibrium = 0.17 mol

Calculate Partial Pressures

The total pressure is 0.25 atm. We will use the mole fraction to find partial pressures:\[x_{NOBr} = \frac{0.66}{0.66 + 0.68 + 0.17} = \frac{0.66}{1.51} = 0.437 \quad \Rightarrow \quad P_{NOBr} = 0.437 \times 0.25 = 0.109 \text{ atm}\]\[x_{NO} = \frac{0.68}{1.51} = 0.450 \quad \Rightarrow \quad P_{NO} = 0.450 \times 0.25 = 0.1125 \text{ atm}\]\[x_{Br2} = \frac{0.17}{1.51} = 0.112 \quad \Rightarrow \quad P_{Br2} = 0.112 \times 0.25 = 0.028 \text{ atm}\]

Calculate Equilibrium Constant \(K_P\)

The expression for the equilibrium constant in terms of pressure, \(K_P\), is:\[K_P = \frac{{(P_{NO})^2 (P_{Br2})}}{{(P_{NOBr})^2}} = \frac{{(0.1125)^2 (0.028)}}{{(0.109)^2}}\]Calculating gives:\[K_P = \frac{0.012680625 \times 0.028}{0.011881} = 0.0298 \]

Convert \(K_P\) to \(K_c\)

We use the relationship between \(K_P\) and \(K_c\):\[K_P = K_c (RT)^{\Delta n}\]where \(\Delta n = 3 - 2 = 1\), and \(R = 0.0821 \text{ L atm mol}^{-1} \text{K}^{-1}\). At \(298 \text{ K}\),\[K_c = \frac{K_P}{RT} = \frac{0.0298}{0.0821 \times 298} = 0.001227\]

Conclusion

The equilibrium constants for the dissociation of NOBr at 25°C are:- \(K_P = 0.0298\)- \(K_c = 0.001227\)

Key concepts

Equilibrium Constant

In chemical equilibrium, a reaction reaches a state where the rate of the forward reaction equals the rate of the reverse reaction. The equilibrium constant is a numerical value that represents this balance. For reactions involving gases, the equilibrium constant can be expressed in terms of partial pressures, known as \( K_P \), or in terms of concentrations, known as \( K_c \). Understanding these constants helps us predict the direction of the reaction and the concentrations of reactants and products at equilibrium.

• \( K_P \) is used when dealing with gases where pressure is a practical measure.
• \( K_c \) is used when the reaction components are expressed in terms of molarity.

To convert \( K_P \) to \( K_c \), we can use the equation: \[K_P = K_c (RT)^{\Delta n}\]where \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas, calculated as the number of moles of gaseous products minus the number of moles of gaseous reactants. This conversion is essential for making equilibrium calculations consistent across different measurement systems.

Partial Pressure

In gas-phase reactions, the individual pressures that each gas exerts in a mixture contribute to its total pressure, known as partial pressure. Each gas in a mixture behaves independently, and its partial pressure is proportional to its mole fraction multiplied by the total pressure. It provides a more refined insight when considering chemical equilibria involving gases.Partial pressures can be calculated using the formula:\[P_i = x_i \times P_{\text{total}}\]where:

• \( P_i \) is the partial pressure of component \( i \).
• \( x_i \) is the mole fraction of component \( i \), determined by dividing its moles by the total moles of the gas mixture.
• \( P_{\text{total}} \) is the total pressure of the gas mixture.

For instance, in a reaction, knowing the partial pressure of each component allows us to compute the equilibrium constant \( K_P \). This constant relates the equilibrium concentrations (or pressures) of products and reactants and is vital in predicting how changes in conditions will affect the equilibrium.

Dissociation

Dissociation in chemistry refers to the process in which complex molecules break apart into simpler constituents, such as atoms or ions. In the context of gases, like nitrosyl bromide (NOBr), dissociation signifies the splitting of molecules into simpler gaseous products. For the exercise at hand, 34% dissociation of \( ext{NOBr} \) implies that 34% of its initial moles have broken down into nitrogen oxide (NO) and bromine (Br₂). This concept is crucial in understanding gas-phase reactions and their equilibrium states.Calculating dissociation involves:

• Identifying the initial quantity of the substance.
• Quantifying the percent dissociated, which states what portion of the original substance has reacted.
• Assessing the concentration of products formed from the dissociated molecules.

Dissociation percentages help us calculate the number of moles of each component at equilibrium, which then facilitates the determination of partial pressures and the equilibrium constant. Understanding these transformations is essential for solving equilibrium problems involving gaseous reactions.