Problem 38 - Chemistry and Chemical Reactivity

Question

At 2300 K the equilibrium constant for the formation of \(\mathrm{NO}(\mathrm{g})\) is \(1.7 \times 10^{-3}\) $$ \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) $$ (a) Analysis shows that the concentrations of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) are both \(0.25 \mathrm{M},\) and that of \(\mathrm{NO}\) is \(0.0042 \mathrm{M}\) under certain conditions. Is the system at equilibrium? (b) If the system is not at equilibrium, in which direction does the reaction proceed? (c) When the system is at equilibrium, what are the equilibrium concentrations?

Step-by-Step Solution

Answer

(a) System not at equilibrium; \( Q < K_c \). (b) Reaction proceeds to the right. (c) Use \( K_c \) and ICE table to find concentrations.

Step 1: Write the Expression for the Reaction Quotient (Q)

The reaction quotient \( Q \) is written in the same way as the equilibrium constant \( K_c \), which is: \[ Q = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} \] Substitute the given concentrations into this expression: \[ Q = \frac{(0.0042)^2}{(0.25)(0.25)} \]

Step 2: Calculate the Reaction Quotient (Q)

Calculate \( Q \) using the expression from Step 1: \[ Q = \frac{0.00001764}{0.0625} = 0.00028224 \]

Step 3: Compare Q with the Equilibrium Constant (K)

By comparing \( Q = 0.00028224 \) with the given \( K_c = 1.7 \times 10^{-3} \), we find that \( Q < K_c \). This means the system is not at equilibrium.

Step 4: Determine the Direction of the Reaction

Since \( Q < K_c \), the system will shift to the right to reach equilibrium, which means the forward reaction (forming more \( \text{NO} \)) will proceed.

Step 5: Set Up the ICE Table

Let the change in concentration for \( \text{NO} \) be \( +2x \). Then for \( \text{N}_2 \) and \( \text{O}_2 \), it must be \( -x \) for each, due to the stoichiometry of the reaction. Set up as follows:| | \( \text{N}_{2} \) | \( \text{O}_{2} \) | \( \text{NO} \) ||------|-------------|-------------|---------|| I | 0.25 | 0.25 | 0.0042 || C | -x | -x | +2x || E | 0.25-x | 0.25-x | 0.0042+2x |

Step 6: Use the Kc Expression to Solve for x

Use the equilibrium expression \( K_c = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} \) and solve for \( x \):\[ 1.7 \times 10^{-3} = \frac{(0.0042+2x)^2}{(0.25-x)(0.25-x)} \]Calculate \( x \) algebraically.

Step 7: Calculate the Equilibrium Concentrations

Once \( x \) is determined, substitute it back into the expressions for the equilibrium concentrations:- \([\text{N}_2] = 0.25 - x\)- \([\text{O}_2] = 0.25 - x\)- \([\text{NO}] = 0.0042 + 2x\)Use the value of \( x \) to calculate these concentrations.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a fundamental concept in chemical equilibrium. It represents the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. For the reaction \( \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2\mathrm{NO}(\mathrm{g}) \), the equilibrium constant at \( 2300 \text{ K} \) is given as \( 1.7 \times 10^{-3} \). This small value of \( K_c \) indicates that at equilibrium, the concentrations of \( \mathrm{NO} \) are much lower compared to \( \mathrm{N}_2 \) and \( \mathrm{O}_2 \), signifying that the reactants are favored under these conditions.
It is important to remember that the equilibrium constant is temperature-dependent, meaning that any change in temperature will affect the value of \( K_c \). However, \( K_c \) remains constant at a given temperature regardless of the initial concentrations of reactants and products. Understanding this helps in predicting the direction of reaction based on comparing \( K_c \) with another key concept, the reaction quotient \( Q \).

Reaction Quotient

The reaction quotient \( Q \) is a mathematical expression similar to the equilibrium constant \( K_c \), which can be calculated using the current concentrations of reactants and products.
It is expressed as:

• \( Q = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} \).

For this exercise, substituting the given concentrations in the formula resulted in \( Q = 0.00028224 \). The purpose of calculating \( Q \) is to determine the state of the system in comparison to its equilibrium state.

By comparing \( Q \) to \( K_c \):

• If \( Q = K_c \), the system is at equilibrium.
• If \( Q < K_c \), the forward reaction is favored to reach equilibrium, meaning more products will form.
• If \( Q > K_c \), the reverse reaction is favored to reach equilibrium, leading to the formation of more reactants.

In this case, since \( Q < K_c \), the system will favor the forward reaction, resulting in the production of more \( \mathrm{NO} \).

Le Chatelier's Principle

Le Chatelier's Principle is an essential concept in understanding how a system at equilibrium responds to disturbances or changes in conditions.
It states that if a dynamic equilibrium is disturbed by changing the conditions, such as concentration, temperature, or pressure, the position of equilibrium shifts to counteract the change and re-establish equilibrium. For this reaction, since \( Q < K_c \), it indicates deviation from equilibrium, triggering a response to minimize this disturbance.

Applying Le Chatelier's Principle:

• If more reactants \( (\text{N}_2 \text{ or } \text{O}_2) \) are added, the system will shift to the right, forming more \( \text{NO} \) to re-balance.
• If product \( (\text{NO}) \) is removed, the system similarly shifts right, producing more \( \text{NO} \).
• A decrease in temperature (for exothermic reactions) or increase in pressure (favoring fewer gas molecules) could also shift equilibrium.

Understanding this helps in both predicting the reaction's behavior under new conditions and designing experiments to manipulate the outcome.

Equilibrium Concentrations

Equilibrium concentrations refer to the concentrations of reactants and products in a reaction when the system has reached equilibrium.
In the context of this exercise, finding the equilibrium concentrations involves using the ICE (Initial, Change, Equilibrium) table method. This method is helpful for predicting how concentrations change with time until equilibrium is re-established. An ICE table accounts for initial concentrations (I), changes in concentration (C), and equilibrium concentrations (E).

Once the value of \( x \) (the change in concentration for \( \mathrm{NO} \) is determined through solving the equilibrium expression:

• \([\mathrm{NO}] = 0.0042 + 2x \)
• \([\mathrm{N}_2] = 0.25 - x \)
• \([\mathrm{O}_2] = 0.25 - x \)

Substituting \( x \) back into these expressions will give the equilibrium concentrations for each species. These values are foundational for understanding how various factors affect a chemical reaction in reaching equilibrium.