Problem 22 - Chemistry: The Central Science

Question

Consider the following equilibrium, for which \(K_{p}=0.0752\) at \(480^{\circ} \mathrm{C}\) $$ 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) $$ (a) What is the value of \(K_{p}\) for the reaction \(4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) ?\) (b) What is the value of \(K_{p}\) for the reaction \(\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?\) (c) What is the value of \(K_{c}\) for the reaction in part (b)?

Step-by-Step Solution

Answer

The following equilibrium constants are found: (a) For the reverse reaction, K_p ≈ 13.3. (b) For the halved coefficients reaction, K_p ≈ 0.274. (c) For the halved coefficients reaction, K_c ≈ 6.81 x 10^-3.

Step 1: In order to find the K_p value for the reverse reaction, which is the inverse of the original reaction, simply take the reciprocal of the original K_p value. Reverse K_p = \(\frac{1}{K_{p}}\) Put the value of K_p (0.0752) in the above formula. #Step 2: Calculate the reverse K_p#

Reverse K_p = \(\frac{1}{0.0752}\) ≈ 13.3 So, the K_p value for the reverse reaction is about 13.3. #Step 3: Find K_p for the halved coefficients reaction (b)#

Step 2: To determine the K_p for the reaction with halved coefficients, raise the original K_p to a power that equals the ratio of the coefficients (1/2). K_p (halved) = \((K_{p_{original}})^{\frac{1}{2}}\) Substitute the value of K_p (0.0752) into the formula. #Step 4: Calculate the K_p for the halved coefficients reaction#

K_p (halved) = \((0.0752)^{\frac{1}{2}}\) ≈ 0.274 So, the K_p value for the reaction with halved coefficients is about 0.274. #Step 5: Find the relationship between K_p and K_c (c)#

Step 3: The relationship between K_p and K_c is given by the equation: \(K_p = K_c(RT)^{\Delta n}\) K_p = equilibrium constant in terms of pressure K_c = equilibrium constant in terms of concentration R = ideal gas constant (0.0821 L atm/mol K) T = temperature in Kelvin (480 + 273.15 = 753.15 K) Δn = change in the number of moles of gas (products - reactants) For the reaction in part (b): Δn = (2 + 1/2) - (1 + 1) = 1/2 #Step 6: Calculate the K_c value for the reaction in part (b)#

Use the given relationship formula, plug in the values, and solve for K_c: \(K_c = \frac{K_p}{(RT)^{\Delta n}}\) \(K_c = \frac{0.274}{(0.0821 \cdot 753.15)^{\frac{1}{2}}}\) K_c ≈ 6.81 x 10^-3 So, the K_c value for the reaction in part (b) is about 6.81 x 10^-3.

Key Concepts

Equilibrium Constant Kp

The Equilibrium Constant, denoted as \( K_p \), is crucial when reactions involve gases, and it is related to the pressures of the reactants and products. To find the \( K_p \) for a reaction, you use the partial pressures of each gas species at equilibrium. The equation depends on the stoichiometry of the reaction, essentially the balance of how much product and reactant gas "exert" pressure in the context of the reaction.
For example, given the reaction:

• \(2 \text{Cl}_2(g) + 2 \text{H}_2\text{O}(g) \rightleftharpoons 4 \text{HCl}(g) + \text{O}_2(g)\)

The \( K_p \) expression for this reaction would be:

• \[K_p = \frac{(P_{\text{HCl}})^4(P_{\text{O}_2})}{(P_{\text{Cl}_2})^2(P_{\text{H}_2\text{O}})^2}\]

Reversing this reaction or changing coefficients also involves mathematical adjustments to \( K_p \). For reversal, use the reciprocal. For coefficient changes, adjust by using powers.

Equilibrium Constant Kc

Unlike \( K_p \), the Equilibrium Constant \( K_c \) refers to concentration-based calculations instead of pressure. Concentrations are typically measured in molarity (moles per liter), and \( K_c \) gives insight into the relative concentrations of products and reactants at equilibrium. Each form of constant provides useful insights depending on reaction conditions.
To determine \( K_c \), particularly when given \( K_p \), you can transform between these constants by using the equation:

• \(K_p = K_c (RT)^{\Delta n}\)

Where:

• \( R \) is the ideal gas constant, \( 0.0821\ \text{L atm/mol K} \)
• \( T \) is the temperature in Kelvin
• \( \Delta n \) is the change in moles of gas from reactants to products.

This relationship indicates how temperature and mole changes influence the equilibrium constants' equivalence. Adjustments based on \( \Delta n \) accommodate the phase shift from concentration to pressure.

Reaction Quotients

The Reaction Quotient, \( Q \), has a similar formulation to \( K \) (either \( K_c \) or \( K_p \)), but \( Q \) serves a different purpose—it helps determine the direction the reaction will proceed to reach equilibrium. By comparing \( Q \) to \( K \), you can predict if a reaction is at equilibrium or if it'll shift towards reactants or products.
Here's a brief rundown:

• If \( Q < K \), the reaction shifts to the right, forming more products.
• If \( Q > K \), the reaction shifts to the left, producing more reactants.
• If \( Q = K \), the system is at equilibrium.

This is practical when starting with non-equilibrium concentrations or pressures to anticipate how the system will adjust over time.

Ideal Gas Law

The Ideal Gas Law is fundamental in equilibrium calculations involving gases. Its standard form is \( PV = nRT \), linking pressure \( P \), volume \( V \), moles \( n \), the ideal gas constant \( R \), and temperature \( T \).
This law is crucial when discussing \( K_p \) because it allows us to switch between moles and pressures, which is especially relevant when we need to convert conditions into a form suitable for equilibrium expression calculations. We often use this principle to derive pressure given other conditions, aiding in accurate \( K_p \) assessments.
In equilibrium contexts:

• Used to convert concentrations (from \( n/V \)) into pressures for gaseous equilibria calculations.
• Remember, \( R = 0.0821\ \text{L atm/mol K} \) keeps units consistent when temperature is in Kelvin.

Ultimately, the Ideal Gas Law provides the scaffolding that ties physical states to equilibrium expressions, ensuring accurate and coherent calculations.