Question

Nitricoxide (NO) reacts readily with chlorine gas as follows: $$ 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g) $$ At \(700 \mathrm{~K}\) the equilibrium constant \(K_{p}\) for this reaction is \(0.26\). Predict the behavior of each of the following mixtures at this temperature: (a) \(P_{\mathrm{NO}}=0.15 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.31 \mathrm{~atm}\), and \(P_{\mathrm{NOC} 1}=0.11 \mathrm{~atm} ;\) (b) \(\mathrm{P}_{\mathrm{NO}}=0.12 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.10 \mathrm{~atm}\) and \(\quad P_{\text {NOCl }}=0.050\) atm; \(\quad\) (c) \(\quad P_{N O}=0.15\) atm, \(P_{\mathrm{C}_{2}}=0.20 \mathrm{~atm}\), and \(P_{\mathrm{NOCl}}=5.10 \times 10^{-3} \mathrm{~atm}\)

Short answer

For all three mixtures, the reaction will proceed in the forward direction to reach the equilibrium, as their calculated Qp values (0.171, 0.174, and 1.16 x 10^{-3}) are less than the given Kp value (0.26). This means more NOCl will be formed in each case to reach equilibrium at 700 K.

Step-by-step solution

Write down the balanced chemical equation and equilibrium constant expression.

The balanced chemical equation for the reaction is: \[2NO(g) + Cl_2(g) \rightleftharpoons 2NOCl(g)\] The expression for the equilibrium constant, Kp, in terms of partial pressures is: \[K_p = \frac{P_{\mathrm{NOCl}}^2}{P_{\mathrm{NO}}^2 P_{\mathrm{Cl}_2}}\]

Calculate Qp for each mixture

Calculate the initial reaction quotient, Qp, for all three mixtures using the provided partial pressures. (a) P_NO = 0.15 atm, P_Cl2 = 0.31 atm, P_NOCl = 0.11 atm \[Q_p = \frac{P_{\mathrm{NOCl}}^2}{P_{\mathrm{NO}}^2 P_{\mathrm{Cl}_2}} = \frac{(0.11)^2}{(0.15)^2(0.31)} = 0.171\] (b) P_NO = 0.12 atm, P_Cl2 = 0.10 atm, P_NOCl = 0.050 atm \[Q_p = \frac{P_{\mathrm{NOCl}}^2}{P_{\mathrm{NO}}^2 P_{\mathrm{Cl}_2}} = \frac{(0.050)^2}{(0.12)^2(0.10)} = 0.174\] (c) P_NO = 0.15 atm, P_Cl2 = 0.20 atm, P_NOCl = 5.10 x 10^{-3} atm \[Q_p = \frac{P_{\mathrm{NOCl}}^2}{P_{\mathrm{NO}}^2 P_{\mathrm{Cl}_2}} = \frac{(5.10 \times 10^{-3})^2}{(0.15)^2(0.20)} = 1.16\times10^{-3}\]

Compare Qp and Kp and predict the behavior

For each mixture, compare the calculated Qp with the given Kp (0.26) and predict whether the reaction will proceed in the forward or reverse direction to reach equilibrium. (a) Qp (0.171) < Kp (0.26) The reaction will proceed in the forward direction (more NOCl will be formed) to reach equilibrium. (b) Qp (0.174) < Kp (0.26) The reaction will proceed in the forward direction (more NOCl will be formed) to reach equilibrium. (c) Qp (1.16 x 10^{-3}) < Kp (0.26) The reaction will proceed in the forward direction (more NOCl will be formed) to reach equilibrium. In summary, for all three mixtures, the reaction will proceed in the forward direction to reach the equilibrium.

Key concepts

Reaction Quotient

The reaction quotient, denoted as \(Q_p\), is a crucial concept when discussing chemical equilibrium. It provides a snapshot of a reaction's state at any given moment, allowing us to assess the relative amounts of reactants and products. Importantly, \(Q_p\) is calculated using the same formula as the equilibrium constant \(K_p\), but it is based on the current, not equilibrium, concentrations of reactants and products.

For our given reaction \(2NO(g) + Cl_2(g) \rightleftharpoons 2NOCl(g)\), the expression for \(Q_p\) is:

• \[Q_p = \frac{P_{\mathrm{NOCl}}^2}{P_{\mathrm{NO}}^2 P_{\mathrm{Cl}_2}}\]

The calculated \(Q_p\) helps to predict which way the reaction needs to proceed to reach equilibrium by comparing it to \(K_p\). If \(Q_p < K_p\), the reaction will shift forward, forming more products. Conversely, if \(Q_p > K_p\), it will reverse. When \(Q_p = K_p\), the reaction is at equilibrium.

Equilibrium Constant

The equilibrium constant, \(K_p\), is a fundamental part of understanding chemical equilibria. It defines the ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced equation, when the reaction has reached equilibrium.

For our specific reaction \(2NO(g) + Cl_2(g) \rightleftharpoons 2NOCl(g)\), the equilibrium constant \(K_p\) at 700 K is given as 0.26. The expression is:

• \[K_p = \frac{P_{\mathrm{NOCl}}^2}{P_{\mathrm{NO}}^2 P_{\mathrm{Cl}_2}}\]

This constant is unique to each reaction and depends on the temperature. A \(K_p\) value higher than 1 typically indicates a product-favored equilibrium, while a value lower than 1 indicates a reactant-favored equilibrium.

In our examples, a \(K_p\) of 0.26 tells us that, at this temperature, the equilibrium is somewhat reactant-favored, meaning more parent molecules (NO and \(Cl_2\)) than products (\(NOCl\)) will be present at equilibrium.

Partial Pressure

Partial pressure is an essential aspect when working with gases in chemical reactions. It refers to the pressure exerted by a single type of gas in a mixture of gases. In reactions involving gases, understanding and utilizing partial pressures are crucial in calculating reaction quotients and equilibrium constants.

For the reaction \(2NO(g) + Cl_2(g) \rightleftharpoons 2NOCl(g)\) given in the exercise, the partial pressures of each gas need to be known so that we can compute both \(Q_p\) and \(K_p\).

How to Use Partial Pressures:

• Determine the pressure of each gaseous component separately in the mixture.
• Use these values directly in the expressions for \(Q_p\) and \(K_p\).
• Calculate and compare these to predict the direction of the reaction.

In summary, the partial pressure is both a measure of a gas’s concentration within a mixture and a tool that informs the behavior of the reaction components.

Predicting Reaction Direction

One of the primary uses of \(Q_p\) and \(K_p\) is to predict the direction in which a reaction will naturally proceed to reach equilibrium. This process involves comparing \(Q_p\) against \(K_p\).

Guidelines for Prediction:

• If \(Q_p < K_p\): The reaction will move in the forward direction to produce more products until equilibrium is achieved.
• If \(Q_p > K_p\): The reaction will shift in the reverse direction, converting products back into reactants.
• If \(Q_p = K_p\): The reaction is considered to be at equilibrium; no net change will occur.

For all the examples given (a, b, and c), where \(Q_p\) was calculated to be less than \(K_p\) (0.26), it's predicted that the reaction will proceed forward. This means more \(NOCl\) will form until the system reaches equilibrium.