Problem 126 - Chemistry

Question

Given \(K=3.50\) at \(45^{\circ} \mathrm{C}\) for the reaction $$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)$$ and \(K=7.10\) at \(45^{\circ} \mathrm{C}\) for the reaction $$2 \mathrm{A}(g)+\mathrm{D}(g) \rightleftharpoons \mathrm{C}(g)$$ what is the value of \(K\) at the same temperature for the reaction $$\mathrm{C}(g)+\mathrm{D}(g) \rightleftharpoons 2 \mathrm{B}(g)$$ What is the value of \(K_{\mathrm{p}}\) at \(45^{\circ} \mathrm{C}\) for the reaction? Starting with 1.50 atm partial pressures of both \(\mathrm{C}\) and \(\mathrm{D},\) what is the mole fraction of \(\mathrm{B}\) once equilibrium is reached?

Step-by-Step Solution

Answer

The value of \(K_3\) at the same temperature for the reaction \(C(g) + D(g) \rightleftharpoons 2B(g)\) is 1.725. The mole fraction of B once equilibrium is reached is approximately 0.312.

Step 1: Write down all given information

We are given the following information: 1. For the reaction: \(A(g) + B(g) \rightleftharpoons C(g)\), we have \(K_1 = 3.50\) at \(45^\circ \mathrm{C}\) 2. For the reaction: \(2A(g) + D(g) \rightleftharpoons C(g)\), we have \(K_2 = 7.10\) at \(45^\circ \mathrm{C}\) 3. We need to find \(K_3\), the equilibrium constant for the reaction: \(C(g) + D(g) \rightleftharpoons 2B(g)\)

Step 2: Manipulate the given reactions to obtain the target reaction

First, we will multiply the first reaction by 2, to make the number of A's equal to the second reaction, resulting in: \[2A(g) + 2B(g) \rightleftharpoons 2C(g) \quad (1')\] Now, we can subtract the second reaction from the first reaction (1'): \((1') - (2) \Rightarrow )[2A(g) + 2B(g) \rightleftharpoons 2C(g)] - [2A(g) + D(g) \rightleftharpoons C(g)]\] Which results in the target reaction: \[C(g) + D(g) \rightleftharpoons 2B(g)\]

Step 3: Apply the rule for equilibrium constants of the manipulated reactions and find the \(K_3\)

When we multiplied the first reaction by 2, the new equilibrium constant becomes \(K_1^{'} = K_1^2\), so we get: \(K_1^{'} = 3.50^2 = 12.25\) When we subtracted the second reaction from the first reaction, the new equilibrium constant will be the quotient of the two constants \(K_3 = \frac{K_1^{'}}{K_2}\), so we have: \(K_3 = \frac{12.25}{7.10} = 1.725\) So, the value of \(K_3\) at the same temperature for the reaction \(C(g) + D(g) \rightleftharpoons 2B(g)\) is 1.725.

Step 4: Calculate the mole fraction of B at equilibrium

We are given that the initial partial pressures of C and D are both 1.50 atm. Let the change in pressure of B at equilibrium be denoted as \(x\). Then, the pressure of B at equilibrium will be \(2x\), and the pressure of C and D will be \(1.50 - x\) atm each. The reaction quotient, \(Q_p\), for the target reaction can be written as: \(Q_p = \frac{P^2_{B}}{P_{C} \cdot P_{D}}\) At equilibrium, \(Q_p\) is equal to \(K_p\). Thus, \(K_p = \frac{(2x)^2}{(1.50 - x)(1.50 - x)}\) We are given that \(K_p = K_3 = 1.725\), so: \(1.725 = \frac{4x^2}{(1.50 - x)^2}\) Now, we will solve for \(x\). Since we are looking for the mole fraction of B, which must be less than 1, we can discard any extraneous solutions. \(x \approx 0.468\) The mole fraction of B at equilibrium is: \[\frac{2x}{1.50 + 1.50} = \frac{2(0.468)}{3.00} \approx 0.312\] Thus, the mole fraction of B once equilibrium is reached is approximately 0.312.

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to the state in which the concentrations of reactants and products in a chemical reaction remain constant over time.
This happens when the forward and reverse reactions occur at the same rate. In an equilibrium state, the reaction doesn't stop; both directions continue to proceed, but there's no net change in the concentrations. Equilibrium can be reached in both closed and open systems.
For a reaction like \[ A(g) + B(g) \rightleftharpoons C(g), \]if the system reaches equilibrium, the rate at which A and B form C is equal to the rate at which C decomposes back into A and B.

• The equilibrium constant, \( K \), gives us the ratio of the concentrations of products to reactants at equilibrium, allowing us to understand the extent of a reaction under given conditions.

Reaction Quotient

The reaction quotient \( Q \) is a tool used to determine the direction in which a chemical reaction will proceed.
It's similar to the equilibrium constant \( K \) but can be calculated at any point in time, not just at equilibrium.
To find \( Q \), use the same formula as \( K \), substituting the current concentrations or pressures of the reactants and products. If a reaction is in the form:\[ C(g) + D(g) \rightleftharpoons 2B(g), \]you calculate \( Q \) as:\[ Q = \frac{P_B^2}{P_C \cdot P_D}. \]

• If \( Q > K \), the reaction will shift to the left to reach equilibrium.
• If \( Q < K \), the reaction will shift to the right.
• If \( Q = K \), the system is at equilibrium.

This tool is essential when predicting how a system will behave when conditions change.

Partial Pressures

Partial pressures refer to the pressure contributed by each individual gas in a mixture.
In a chemical reaction involving gases, each gas behaves independently and contributes to the total pressure.
This concept is crucial when working with equilibrium calculations involving gases. For a mixture containing gases C, D, and B, partial pressures are used to calculate the equilibrium constant expression.
If initial partial pressures are given, like 1.50 atm for both C and D, they help in setting up equations to find changes in pressure during the reaction.
For the reaction: \[ C(g) + D(g) \rightleftharpoons 2B(g), \]the partial pressures of each gas allow us to solve for the equilibrium position using: \[ K_p = \frac{P_B^2}{P_C \cdot P_D}, \]where \( P_B, P_C, \) and \( P_D \) are the equilibrium pressures of B, C, and D, respectively.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture.
It is defined as the ratio of the moles of one component to the total moles in the mixture.
In a chemical equilibrium problem, understanding mole fractions helps determine concentrations of gases relative to each other. For the reaction \[ C(g) + D(g) \rightleftharpoons 2B(g), \]if the equilibrium condition changes the pressures of C and D from 1.50 atm to 1.50 - x atm, and the pressure of B to 2x atm, the mole fraction of B is given by:

• \( \text{Mole fraction of } B = \frac{2x}{1.50 + 1.50} \).

This reflects B's proportion in the total pressure of the system. Mole fractions are dimensionless and valuable in determining reaction outcomes.