Question

At \(1000 \mathrm{~K}\), a sample of pure \(\mathrm{NO}_{2}\) gas decomposes as: $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The equilibrium constant \(K_{p}\) is \(156.25\) atm. Analysis shows that the partial pressure of \(\mathrm{O}_{2}\) is \(0.25\) atm at equilibrium. The partial pressure of \(\mathrm{NO}_{2}\) at equilibrium is : (a) \(0.01\) (b) \(0.02\) (c) \(0.04\) (d) None of these

Short answer

The partial pressure of NO2 at equilibrium is 0.04 atm, hence the correct answer is (c).

Step-by-step solution

Write down the equilibrium expression for Kp

Based on the decomposition reaction of NO2, the equilibrium constant expression for the reaction in terms of partial pressures (Kp) is: \[ K_{p} = \frac{(P_{\mathrm{NO}})^2 \times P_{\mathrm{O}_2}}{(P_{\mathrm{NO}_2})^2} \] Where:\[ P_{\mathrm{NO}} \] is the partial pressure of NO,\[ P_{\mathrm{O}_2} \] is the partial pressure of O2, and \[ P_{\mathrm{NO}_2} \] is the partial pressure of NO2 at equilibrium.

Determine the stoichiometry of the reaction

From the balanced chemical equation, we can see that 2 moles of NO2 decompose to form 2 moles of NO and 1 mole of O2. Therefore, we can express the moles of NO formed in terms of the moles of NO2 decomposed. Let the decrease in moles of NO2 (and hence the partial pressure of NO2) be x. Then the increase in moles (and partial pressure) of NO will be 2x (since 2 moles of NO are formed for every 2 moles of NO2 decomposed) and the increase in partial pressure of O2 will be x (since 1 mole of O2 is formed for every 2 moles of NO2 decomposed).

Relate the given equilibrium partial pressure of O2 to Kp

The partial pressure of O2 at equilibrium is given as 0.25 atm. Hence, x, the change in partial pressure of NO2, is 0.25 atm. We can use this value and the value of Kp to find the equilibrium partial pressure of NO2. Substitute the known values into the equilibrium expression: \[ 156.25 = \frac{(2 \times 0.25)^2}{(P_{\mathrm{NO}_2})^2} \]

Solve for the partial pressure of NO2

Resolve the expression to find the value of \(P_{\mathrm{NO}_2}\): \[ 156.25 = \frac{(0.5)^2}{(P_{\mathrm{NO}_2})^2} \] \[ 156.25 = \frac{0.25}{(P_{\mathrm{NO}_2})^2} \] \[ (P_{\mathrm{NO}_2})^2 = \frac{0.25}{156.25} \] \[ (P_{\mathrm{NO}_2})^2 = \frac{1}{625} \] \[ P_{\mathrm{NO}_2} = \frac{1}{25} \] \[ P_{\mathrm{NO}_2} = 0.04 \text{ atm} \]

Key concepts

Equilibrium Constant

The equilibrium constant, symbolized as K or Kp when using partial pressures, is a quantitative measure of the position of the equilibrium in a chemical reaction. It provides us with a snapshot of the relative concentrations or pressures of reactants and products at equilibrium conditions.

For the reaction at hand:
2NO₂(g) ⇌ 2NO(g) + O₂(g)
The equilibrium constant expression in terms of partial pressures, Kp, is $$K_{p} = \frac{(P_{\text{NO}})^2 \times P_{\text{O}_2}}{(P_{\text{NO}_2})^2}$$Here, the constant is a fixed value at a given temperature (1000 K in the exercise), which in this case is 156.25 atm. It indicates the extent of the reaction; a higher value suggests products are favored, while a lower value suggests reactants are favored. The magnitude of Kp is crucial for predicting the direction of the reaction under different conditions.

Partial Pressure

Partial pressure refers to the pressure that a single gas in a mixture would exert if it were alone in the entire volume of the mixture. This concept is vital when dealing with gas-phase reactions at equilibrium. For each species in a reaction, their respective partial pressure is denoted by P with a subscript of the compound's formula.

In the context of the given reaction, the partial pressure of NO₂ and its relation to NO and O₂ at equilibrium is essential to finding the unknown values. The term 'partial' signifies that it's only part of the total pressure exerted by all gases present.

Equilibrium Expression

The equilibrium expression is a formula that allows us to relate the concentrations or partial pressures of reactants and products at equilibrium for a particular reaction. Setting up the correct expression is crucial to calculate K or Kp.

In our solved exercise, the equilibrium expression was written as a function of the partial pressures of the gases involved, considering the stoichiometry of the balanced equation. Recognizing that each NO₂ molecule that decomposes produces both NO and O₂ simplifies calculations. The exercise successfully demonstrated how to manipulate the expression using the known value of Kp and the given partial pressures to solve for unknowns.