Question

At \(2300 \mathrm{~K}\) 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 of the contents of a sealed flask at \(2300 \mathrm{~K}\) shows that the concentrations of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) are both \(0.25 \mathrm{M}\) and that of \(\mathrm{NO}\) is \(0.0042 \mathrm{M}\). Determine if the system is at equilibrium. (b) If the system is not at equilibrium, in which direction does the reaction proceed? (c) Calculate all three equilibrium concentrations.

Short answer

(a) The system is not at equilibrium. (b) Reaction proceeds forward. (c) Use ICE table to find equilibrium concentrations.

Step-by-step solution

Calculate the Reaction Quotient (Q)

The reaction quotient \( Q \) is given by the expression: \[ Q = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} \] Substituting the given concentrations: \[ Q = \frac{(0.0042)^2}{(0.25)(0.25)} = \frac{0.00001764}{0.0625} = 0.00028224 \]

Compare Q with the Equilibrium Constant (K)

The equilibrium constant \( K \) for the reaction at \( 2300 \mathrm{~K} \) is \( 1.7 \times 10^{-3} \). Compare this value with \( Q \):\[ Q = 0.00028224 \]\[ K = 1.7 \times 10^{-3} \]Since \( Q < K \), the reaction is not at equilibrium and will proceed in the forward direction to increase \([\text{NO}]\).

Set Up an ICE Table

An ICE table helps us determine equilibrium concentrations. Let \( x \) be the change in concentration as the reaction moves toward equilibrium:\[\begin{array}{c|ccc} & \text{N}_2 & \text{O}_2 & \text{NO} \\hline\text{Initial} & 0.25 & 0.25 & 0.0042 \\text{Change} & -x & -x & +2x \\text{Equilibrium} & 0.25-x & 0.25-x & 0.0042+2x \\end{array}\]

Solve for x Using the Equilibrium Constant Equation

Substitute into the equilibrium constant expression:\[ K = \frac{(0.0042 + 2x)^2}{(0.25 - x)(0.25 - x)} = 1.7 \times 10^{-3} \]This simplifies to the equation \((0.25 - x)^2 \times 1.7 \times 10^{-3} = (0.0042 + 2x)^2\). Solve this quadratic equation for \( x \).

Find the Equilibrium Concentrations

Once \( x \) is found from the quadratic equation, substitute back:- \([\text{N}_2] = 0.25 - x\)- \([\text{O}_2] = 0.25 - x\)- \([\text{NO}] = 0.0042 + 2x\)Calculate these to find the equilibrium concentrations of all species.

Key concepts

Reaction Quotient (Q)

The reaction quotient, denoted as \( Q \), is a vital tool for analyzing the state of a chemical reaction at a particular moment in time. It assists in determining whether a system is at equilibrium or if it needs to adjust concentrations to reach equilibrium.

To compute \( Q \), you use the concentrations of the reactants and products at any given point. For our reaction:\[ Q = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} \]Substitute the concentrations from the problem into this expression. Here, \( [\text{NO}] = 0.0042 \text{ M}, [\text{N}_2] = 0.25 \text{ M}, \) and \( [\text{O}_2] = 0.25 \text{ M} \).

Thus, \( Q \) becomes \( 0.00028224 \). Comparing \( Q \) with the equilibrium constant \( K \) reveals the direction the reaction must proceed to reach equilibrium.
- If \( Q < K \), the reaction will proceed in the forward direction.
- If \( Q > K \), it will proceed in the reverse direction.
- If \( Q = K \), the reaction is in equilibrium.

Equilibrium Constant (K)

The equilibrium constant, symbolized as \( K \), is a numerical value that characterizes the ratio of concentrations of products to reactants of a reaction at equilibrium. It's specific to a certain temperature. For instance, in our problem at \( 2300 \text{ K} \), \( K = 1.7 \times 10^{-3} \).

The equilibrium constant is crucial because it allows us to predict the position of equilibrium by comparing it with the reaction quotient \( Q \). Here's how it works:- **High \( K \) Value**: Indicates a propensity for products to be favored over reactants at equilibrium.- **Low \( K \) Value**: Implies that reactants are preferred over products at equilibrium.In the provided scenario, since \( Q < K \), more \( \text{NO} \) needs to form for the system to reach equilibrium. It signals the reaction will proceed forward.

This process of comparing \( Q \) and \( K \) helps predict changes in concentration and determine how far a reaction must go to achieve equilibrium.

ICE Table

An ICE (Initial, Change, Equilibrium) table is a systematic tool used for calculating the equilibrium concentrations of reactants and products in a chemical reaction. It stands for:

• **Initial**: The starting concentrations of reactants and products.
• **Change**: How much the concentrations change as the system moves toward equilibrium.
• **Equilibrium**: The concentrations present once equilibrium is reached.

In the given exercise, the ICE table was set up for the reaction \( \text{N}_2 + \text{O}_2 \rightleftharpoons 2 \text{NO} \). Supposing \( x \) as the change in concentration, it helps visualize the shift in direction:

• **Initial**: \( [\text{N}_2] = 0.25 \text{ M}, [\text{O}_2] = 0.25 \text{ M}, [\text{NO}] = 0.0042 \text{ M} \)
• **Change**: \( -x \) for \( \text{N}_2 \) and \( \text{O}_2 \); \( +2x \) for \( \text{NO} \)
• **Equilibrium**: \( [\text{N}_2] = 0.25 - x, [\text{O}_2] = 0.25 - x, [\text{NO}] = 0.0042 + 2x \)

Using the ICE table, one incorporates the reaction changes to solve the equilibrium expression:\[ K = \frac{(0.0042 + 2x)^2}{(0.25 - x)(0.25 - x)} \]Solving this will give the value of \( x \), which can be substituted back into the expressions for the equilibrium concentrations of each species. It's a practical approach that simplifies handling the complexities of chemical equilibria.