Problem 4 - Chemistry and Chemical Reactivity

Question

The reaction $$ 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftarrows \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) $$ has an equilibrium constant, \(K\), of 170 at \(25^{\circ} \mathrm{C}\). If \(2.0 \times\) \(10^{-3}\) mol of \(\mathrm{NO}_{2}\) is present in a \(10 .-\) L. flask along with \(1.5 \times 10^{-3} \mathrm{mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{4},\) is the system at equilibrium? If it is not at equilibrium, does the concentration of \(\mathrm{NO}_{2}\) increase or decrease as the system proceeds to equilibrium?

Step-by-Step Solution

Answer

No, the system is not at equilibrium. The concentration of \( \text{NO}_2 \) will increase as the system shifts to equilibrium.

Step 1: Write the Expression for the Equilibrium Constant

The equilibrium constant \( K \) for the reaction \( 2 \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) \) is given by:\[ K = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2} \] where \([\text{N}_2\text{O}_4]\) and \([\text{NO}_2]\) are the concentrations of \(\text{N}_2\text{O}_4\) and \(\text{NO}_2\) at equilibrium.

Step 2: Calculate Initial Concentrations

Convert moles to concentrations using the volume of the container. - Concentration of \(\text{NO}_2\): \[ [\text{NO}_2] = \frac{2.0 \times 10^{-3} \text{ mol}}{10 \text{ L}} = 2.0 \times 10^{-4} \text{ M} \]- Concentration of \(\text{N}_2\text{O}_4\): \[ [\text{N}_2\text{O}_4] = \frac{1.5 \times 10^{-3} \text{ mol}}{10 \text{ L}} = 1.5 \times 10^{-4} \text{ M} \]

Step 3: Calculate the Reaction Quotient, Q

The reaction quotient \( Q \) is calculated similarly to \( K \), using initial concentrations:\[ Q = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2} = \frac{1.5 \times 10^{-4}}{(2.0 \times 10^{-4})^2} = \frac{1.5 \times 10^{-4}}{4.0 \times 10^{-8}} = 3750 \]

Step 4: Determine if the System is at Equilibrium

Compare \( Q \) and \( K \):- \( K = 170 \) and \( Q = 3750 \).- Since \( Q > K \), the system is not at equilibrium and the reaction will shift to the left to reach equilibrium.

Step 5: Predict Changes in Concentrations

Since the reaction will shift to the left, more \( \text{N}_2\text{O}_4 \) will convert into \( \text{NO}_2 \), which means the concentration of \( \text{NO}_2 \) will increase as the system proceeds to equilibrium.

Key Concepts

Reaction Quotient

The reaction quotient, represented as \( Q \), is a valuable tool for predicting the direction a chemical reaction will proceed to establish equilibrium. It is calculated in the same way as the equilibrium constant, \( K \), but uses the initial concentrations of the reactants and products instead of those at equilibrium. For the given reaction \( 2 \mathrm{NO}_{2}( ext{g}) \rightleftharpoons \mathrm{N}_{2}\mathrm{O}_{4}( ext{g}) \), \( Q \) is given by the formula:\[ Q = \frac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2} \]In this scenario, we calculated \( Q \) to be 3750 using the initial concentrations. This quotient helps determine if a system is at equilibrium by comparing it to \( K \). If \( Q = K \), the system is at equilibrium. If \( Q eq K \):

• \( Q > K \): The reaction will shift left to produce more reactants.
• \( Q < K \): The reaction will shift right to produce more products.

This understanding helps predict changes in concentration as the reaction achieves equilibrium.

Equilibrium Constant

The equilibrium constant, \( K \), is a vital component of chemical reactions, providing insight into the ratio of concentrations of products to reactants at equilibrium. For the reaction \( 2 \mathrm{NO}_{2}( ext{g}) \rightleftharpoons \mathrm{N}_{2}\mathrm{O}_{4}( ext{g}) \), the equilibrium constant expression is:\[ K = \frac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2} \]Here, \( K \) equals 170 at 25°C. The equilibrium constant is significant because:

• It does not depend on the initial concentrations of reactants and products.
• It changes only with temperature.
• A large \( K \) (much greater than 1) indicates a product-favored reaction at equilibrium.
• A small \( K \) (much less than 1) suggests a reactant-favored reaction.

By comparing \( K \) with \( Q \), one can easily predict in which direction the system will move to reach equilibrium. Understanding \( K \) helps establish what the state of equilibrium looks like for a given reaction.

Le Chatelier's Principle

Le Chatelier's principle is a critical concept in chemical equilibrium, explaining how a system at equilibrium responds to changes or disturbances. According to this principle, if an external change is applied to the system, the equilibrium will shift to counteract the change and restore balance. For the reaction \( 2 \mathrm{NO}_{2}( ext{g}) \rightleftharpoons \mathrm{N}_{2}\mathrm{O}_{4}( ext{g}) \), since \( Q > K \):

• The reaction will shift left, forming more \( \mathrm{NO}_{2} \) and using up \( \mathrm{N_2O_4} \).
• This results in an increase in \( [\mathrm{NO_2}] \) until \( Q \) equals \( K \).

Le Chatelier's principle can also be used to predict the effects of changes in:

• Concentration of reactants or products.
• Pressure and volume for gaseous reactions.
• Temperature changes.

By understanding this principle, one can anticipate how equilibrium shifts under different conditions, allowing for better control of chemical processes.