Problem 60 - Chemistry The Molecular Science

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.

Step-by-Step Solution

Answer

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

Step 1: 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: \[ Q = \frac{(0.0042)^2}{(0.25)(0.25)} = \frac{0.00001764}{0.0625} = 0.00028224 \]

Step 2: 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 \([\text{NO}]\).

Step 3: 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:\[\begin{array}{c|ccc} & \text{N}_2 & \text{O}_2 & \text{NO} \\hline\text{Initial} & 0.25 & 0.25 & 0.0042 \\text{Change} & -x & -x & +2x \\text{Equilibrium} & 0.25-x & 0.25-x & 0.0042+2x \\end{array}\]

Step 4: 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 \( x \).

Step 5: 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.