Question

Consider the reaction $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g) $$ For each of the following mixtures of reactants and products at \(25^{\circ} \mathrm{C}\), predict the direction in which the reaction will shift to reach equilibrium. a. \(P_{\mathrm{NO}_{2}}=P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.0 \mathrm{~atm}\) b. \(P_{\mathrm{NO}_{2}}=0.21 \mathrm{~atm}, P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.50 \mathrm{~atm}\) c. \(P_{\mathrm{NO},}=0.29 \mathrm{~atm}, P_{\mathrm{N}, \mathrm{O}_{4}}=1.6 \mathrm{~atm}\)

Short answer

a. If K > 1, the reaction shifts to the right; if K < 1, the reaction shifts to the left; if K = 1, the system is at equilibrium. b. If K < 11.39, the reaction shifts to the left; if K > 11.39, the reaction shifts to the right; if K = 11.39, the system is at equilibrium. c. If K < 19.07, the reaction shifts to the left; if K > 19.07, the reaction shifts to the right; if K = 19.07, the system is at equilibrium.

Step-by-step solution

Calculate Reaction Quotient (Q)

For any given chemical reaction, we can calculate the reaction quotient (Q) using the formula: $$ Q = \frac{[\textrm{Products}]}{[\textrm{Reactants}]} $$ For the given reaction, $$ Q = \frac{P_{N_{2}O_{4}}}{(P_{NO_{2}})^{2}} $$

Compare Q and K to predict the direction of the reaction

If \(Q < K\), the reaction will shift to the right (favoring formation of products). If \(Q > K\), the reaction will shift to the left (favoring formation of reactants). If \(Q = K\), the system is already at equilibrium. a. \(P_{NO_{2}} = P_{N_{2}O_{4}} = 1.0 \textrm{~atm}\) Calculate Q: $$ Q = \frac{1.0}{(1.0)^{2}} = 1 $$ Now, compare with K (we don't have the numerical value of K, but this is not a problem, as we can still reason about which direction the reaction should shift): Since \(Q = 1\), if K > 1, the reaction will shift to the right (forming more products). If K < 1, the reaction will shift to the left (forming more reactants). If K = 1, the system is already at equilibrium. b. \(P_{NO_{2}} = 0.21 \textrm{~atm}, P_{N_{2}O_{4}} = 0.50 \textrm{~atm}\) Calculate Q: $$ Q = \frac{0.50}{(0.21)^{2}} \approx 11.39 $$ Now, compare with K: Since \(Q > 1\), if K < 11.39, the reaction will shift to the left (forming more reactants). If K > 11.39, the reaction will shift to the right (forming more products). If K = 11.39, the system is already at equilibrium. c. \(P_{NO_{2}} = 0.29 \textrm{~atm}, P_{N_{2}O_{4}} = 1.60 \textrm{~atm}\) Calculate Q: $$ Q = \frac{1.6}{(0.29)^{2}} \approx 19.07 $$ Now, compare with K: Since \(Q > 1\), if K < 19.07, the reaction will shift to the left (forming more reactants). If K > 19.07, the reaction will shift to the right (forming more products). If K = 19.07, the system is already at equilibrium.

Key concepts

Reaction Quotient (Q)

At the heart of understanding chemical equilibrium is the reaction quotient, commonly referred to as Q. It gauges the relative amounts of products and reactants present during a reaction at a particular moment. The formula for the reaction quotient is:

\[\begin{equation}Q = \frac{\text{[Products]}}{\text{[Reactants]}}\end{equation}\]
By using the partial pressures for gaseous systems, we can determine if a reaction at non-equilibrium conditions is leaning towards the products or towards the reactants. When we calculated Q for our specific reaction, we looked at the ratio of the partial pressure of dinitrogen tetroxide (\( P_{N_2O_4} \)) to the square of the partial pressure of nitrogen dioxide (\( (P_{NO_2})^2 \)).

An essential point to note is that for a reaction in a closed system, Q is not static; it changes as the reaction progresses towards equilibrium. Once the system reaches equilibrium, the value of Q becomes equal to the equilibrium constant (K), signifying a balance in the forward and reverse reaction rates.

Le Chatelier's Principle

Le Chatelier's principle is a fantastic tool for predicting how a change in conditions can affect the position of a chemical equilibrium. It states that if a dynamic equilibrium is disturbed by altering the concentration, temperature, or pressure, the system responds by shifting in a direction to counteract the disturbance.

For instance, if we increase the pressure of a gaseous system by decreasing the volume, the equilibrium shifts towards the side with fewer moles of gas. Similarly, increasing the temperature of an exothermic reaction will shift the equilibrium towards the reactants, as the system tries to 'cool down' by absorbing the added heat.

When looking at the problem from our textbook, Le Chatelier's principle guides us to predict the shift in reaction to re-establish equilibrium. Whether it is adding more reactants or products, or changing the pressure or temperature, the reaction shifts accordingly, always striving to maintain balance.

Equilibrium Constant (K)

The equilibrium constant, represented by the symbol K, is a profound expression of the balance point for a chemical reaction at a given temperature. It is calculated using the concentrations of the products and reactants when the reaction has reached its dynamic equilibrium.

Mathematically, the expression for K resembles that of Q but represents the ratios at equilibrium:\[\begin{equation}K = \frac{\text{[Products] equilibrium concentrations}}{\text{[Reactants] equilibrium concentrations}}\end{equation}\]
The magnitude of K gives insight into the favorability of product formation; a large K value signals a strong preference for products at equilibrium, while a small K value indicates a preference for reactants.

When comparing Q and K, the comparison dictates the reaction’s direction to achieve equilibrium. For instance, if Q is less than K, there aren't enough products, and the reaction will proceed to form more. Conversely, if Q is greater than K, there are too many products, and the reaction will shift to reduce them. This fundamental interplay between Q and K underpins all predictive analysis for the behavior of reversible reactions.