Problem 31 - Chemistry

Question

At a particular temperature, a \(3.0-\mathrm{L}\) flask contains 2.4 moles of \(\mathrm{Cl}_{2}, 1.0\) mole of \(\mathrm{NOCl}\) , and \(4.5 \times 10^{-3}\) mole of NO. Calculate \(K\) at this temperature for the following reaction: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$

Step-by-Step Solution

Answer

The equilibrium constant K at this temperature for the given reaction is approximately \(5.74 \times 10^{-3}\).

Step 1: Calculate initial concentrations

We are given the initial moles of all three gases and the volume of the flask, which is 3.0 L. To calculate the initial concentrations, divide the moles by the volume: \([Cl_2] = \frac{2.4 \,\text{moles}}{3.0\,\text{L}} = 0.8\,\text{M}\) \([NOCl] = \frac{1.0 \,\text{moles}}{3.0\,\text{L}} = 0.333\,\text{M}\) \([NO] = \frac{4.5 \times 10^{-3} \,\text{moles}}{3.0\,\text{L}} = 1.5 \times 10^{-3} \,\text{M}\)

Step 2: Calculate the reaction quotient (Q)

Write the expression for Q using the initial concentrations: \(Q = \frac{[NO]^2[Cl_2]}{[NOCl]^2}\) Plug in the initial concentrations for Q: \(Q = \frac{(1.5 \times 10^{-3})^2(0.8)}{(0.333)^2} = 1.80 \times 10^{-5}\)

Step 3: Determine the reaction shift

Since we are asked to calculate the equilibrium constant K, we need to find the direction in which the reaction shifts to reach equilibrium. If Q < K, the reaction will shift to the right (towards products). If Q > K, the reaction will shift to the left (towards reactants). We don't know the value of K yet, but we can still set up the shift expressions for both cases: - If the reaction shifts to the right: \(\Delta NOCl = -2x\), \(\Delta NO = +2x\), and \(\Delta Cl_2 = +x\) - If the reaction shifts to the left: \(\Delta NOCl = +2x\), \(\Delta NO = -2x\), and \(\Delta Cl_2 = -x\)

Step 4: Set up the expressions for final concentrations

Set up the expressions for final concentrations using the shift expressions from step 3: - If the reaction shifts to the right: \([NOCl] = 0.333 - 2x\), \([NO] = 1.5 \times 10^{-3} + 2x\), and \([Cl_2] = 0.8 + x\) - If the reaction shifts to the left: \([NOCl] = 0.333 + 2x\), \([NO] = 1.5 \times 10^{-3} - 2x\), and \([Cl_2] = 0.8 - x\)

Step 5: Find K using the equilibrium concentrations

To find K, rewrite the expressions for final concentrations in terms of Q: - If the reaction shifts to the right: \(K = \frac{[(1.5 \times 10^{-3} + 2x)^2(0.8 + x)]}{(0.333 - 2x)^2} = 1.80 \times 10^{-5}\) - If the reaction shifts to the left: \(K = \frac{[(1.5 \times 10^{-3} - 2x)^2(0.8 - x)]}{(0.333 + 2x)^2} = 1.80 \times 10^{-5}\) Since all concentrations are positive, the reaction must shift to the right. Solving the first equation for x: \(K = \frac{[(1.5 \times 10^{-3} + 2x)^2(0.8 + x)]}{(0.333 - 2x)^2} = 1.80 \times 10^{-5}\) Solving for x gives \(x \approx 1.26 \times 10^{-3}\). Now we can calculate the equilibrium constant K using the final concentrations: \(K = \frac{[(1.5 \times 10^{-3} + 2(1.26 \times 10^{-3}))^2(0.8 + 1.26 \times 10^{-3})]}{(0.333 - 2(1.26 \times 10^{-3}))^2} \approx 5.74 \times 10^{-3}\) Therefore, the equilibrium constant K at this temperature for the given reaction is approximately \(5.74 \times 10^{-3}\).

Key Concepts

Equilibrium Constant (K)

The equilibrium constant, represented as \( K \), is a crucial concept in understanding chemical equilibria. It quantitatively expresses the ratio of concentrations of products to reactants at equilibrium for a given temperature. The value of \( K \) provides insight into the position of equilibrium within a chemical reaction. A large \( K \) indicates that the equilibrium position is towards the products, while a small \( K \) suggests that the equilibrium is towards the reactants. In the exercise, we calculate \( K \) to be approximately \( 5.74 \times 10^{-3} \), implying the system favors the reactants at this temperature. This exercise demonstrates how altering initial concentrations and calculating reaction quotients help us determine equilibrium concentrations, which further leads to the calculation of \( K \). Understanding the equilibrium constant allows us to predict how changes in conditions might affect the balance between reactants and products.

Reaction Quotient (Q)

The reaction quotient, \( Q \), is similar to the equilibrium constant \( K \), but it is calculated using initial concentrations rather than equilibrium concentrations. It helps to determine the direction the reaction will shift in order to reach equilibrium. In the given step-by-step solution, \( Q \) was calculated as \( 1.80 \times 10^{-5} \). By comparing \( Q \) to \( K \), we can infer which way the reaction needs to proceed to achieve equilibrium.
* If \( Q < K \), the reaction will shift to the right, favoring the production of more products.* If \( Q > K \), the reaction will shift to the left, favoring the return towards more reactants.
The computed \( Q \) indicates that it is necessary to assess the reaction's shift based on whether \( Q \) matches \( K \), thereby predicting and confirming the movement of the reaction towards equilibrium.

Concentration Calculations

Concentration calculations are essential for understanding the changes that occur as a reaction moves towards equilibrium. In the exercise, initial concentrations were derived through dividing the number of moles by the volume of the flask (6exvolume \( 3 \text{ L} \)). This calculation is fundamental as it forms the basis for establishing the reaction quotient and ultimately determining equilibrium conditions.
Breaking it down for each compound:

• \([Cl_2]\): Calculated as \(0.8 \text{ M}\)
• \([NOCl]\): Calculated as approximately \(0.333 \text{ M}\)
• \([NO]\): Calculated as \( 1.5 \times 10^{-3} \text{ M} \)

These initial concentrations are critical for calculating \( Q \) and predicting the system's shift to achieve equilibrium. As the reaction progresses, concentration changes can be expressed using variables like \( x \) to represent the change extent, demonstrating the dynamic nature of chemical equilibria.

Le Chatelier's Principle

Le Chatelier's Principle provides insight into how a system at equilibrium responds to disturbances. This principle states that if a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the equilibrium will shift to counteract the change and restore balance.
In the context of our exercise:

• We considered shifts based on initial concentrations and reaction directionality.
• Le Chatelier's Principle helps predict that adding more reactants or products can force the equilibrium position to move, influencing which direction the reaction will shift.

Understanding how these shifts occur under different conditions helps chemists design reactions to favor the desired products or predict how external changes will affect equilibrium states. Overall, Le Chatelier’s Principle is invaluable in predicting the behavior of chemical reactions under various stresses and alterations.