Problem 47 - General Chemistry: Principles and Modern Applications

Question

At \(1000 \mathrm{K},\) an equilibrium mixture in the reaction \(\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \quad\) contains \(0.276 \mathrm{mol} \quad \mathrm{H}_{2}, 0.276 \mathrm{mol} \mathrm{CO}_{2}, \quad 0.224 \mathrm{mol} \mathrm{CO}, \quad\) and \(0.224 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) (a) What is \(K_{\mathrm{p}}\) at \(1000 \mathrm{K} ?\) (b) Calculate \(\Delta G^{\circ}\) at \(1000 \mathrm{K}\). (c) In which direction would a spontaneous reaction occur if the following were brought together at 1000 K: \(0.0750 \mathrm{mol} \mathrm{CO}_{2}, 0.095 \mathrm{mol} \mathrm{H}_{2}, 0.0340 \mathrm{mol} \mathrm{CO},\) and \(0.0650 \mathrm{mol} \mathrm{H}_{2} \mathrm{O} ?\)

Step-by-Step Solution

Answer

\(K_p\) at 1000K is 0.656814, \(\Delta G^\circ\) is 2047.85 J/mol, and the reaction will spontaneously occur towards the products.

Step 1: Preparing for calculating \(K_p\)

To find the equilibrium constant \(K_p\) for the reaction, we first need to notice that we have 4 gases and for gases, a reaction quotient can be expressed in terms of partial pressures. We are given equilibrium number of moles for all gases.

Step 2: Calculate \(K_p\)

The reaction quotient (Q), is defined as \[ Q_p = \frac{P_{CO} P_{H_2O}}{P_{CO2} P_{H2}} \]. On the equilibrium, Q_p becomes the equilibrium constant \(K_p\). The total number of moles is 0.276 + 0.276 + 0.224 + 0.224 = 1. Using ideal gas law, the total pressure for 1 mol at 1000K can be calculated as follows \[ P_{tot}= nRT/V = R*1000/1 =1000R \] and the partial pressure for each gas would be amount of gas in mol times the total pressure. So, \(P_{H2} = \frac{0.276}{1}*1000R = 276R\), \(P_{CO2} = \frac{0.276}{1}*1000R = 276R\), \(P_{CO} = \frac{0.224}{1}*1000R = 224R\) and \(P_{H2O} = \frac{0.224}{1}*1000R = 224R\). Now, calculating \(K_p\) using these values in the equation \[ K_p = \frac{P_{CO} P_{H2O}}{P_{CO2} P_{H2}} = \frac{224R*224R}{276R*276R} = 0.656814 \]

Step 3: Calculate \(\Delta G^\circ\)

Using the relationship between \(K_p\) and \(\Delta G^\circ\) which is \(\Delta G^\circ = -RT \ln K_p\), with R as 8.314 and T at 1000K, we have: \[\Delta G^\circ = -8.314*1000*ln(0.656814) = 2047.85 \ J/mol\]

Step 4: Predict the direction of spontaneous reaction

Here, we again calculate \(Q_p\) using the given number of moles in the reaction mixture using the formula \[ Q_p = \frac{P_{CO} P_{H2O}}{P_{CO2} P_{H2}} \]. After calculating each pressure using the same process as in step 2, we get \(Q_p = 0.294703\). As \(Q_p < K_p\), according to Le Chatelier's principle, the reaction will shift towards right and hence the reaction is spontaneous in the forward direction.

Key Concepts

Reaction Quotient

To understand how reactions progress, we use the reaction quotient, denoted as \( Q \). This nifty tool helps us compare the current state of a reaction mixture to its equilibrium state. For gaseous reactions, we employ the partial pressures of the gases involved. In our case, for the reaction \( \text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g) \), the reaction quotient \( Q_p \) is expressed as: \[ Q_p = \frac{P_{\text{CO}} P_{\text{H}_2\text{O}}}{P_{\text{CO}_2} P_{\text{H}_2}} \].

• If \( Q_p = K_p \), the system is at equilibrium.
• If \( Q_p < K_p \), the reaction will proceed in the forward direction to reach equilibrium.
• If \( Q_p > K_p \), the reaction will proceed in the reverse direction.

By comparing \( Q_p \) to \( K_p \), we determine which way the reaction will "shift" to achieve balance.

Equilibrium Constant

At equilibrium, the rates of the forward and reverse reactions are equal. The equilibrium constant, \( K \), encapsulates the ratio of the concentrations of the products to the reactants at equilibrium. For gaseous reactions, we use \( K_p \), which involves partial pressures instead of concentrations. Calculating \( K_p \) helps predict the favorability of the forward or reverse reaction. The expression for \( K_p \) in our reaction is:\[ K_p = \frac{P_{\text{CO}} P_{\text{H}_2\text{O}}}{P_{\text{CO}_2} P_{\text{H}_2}} \]. At equilibrium, this value informs us about the position of equilibrium:

• A large \( K_p \) indicates that the products are favored at equilibrium.
• A small \( K_p \) suggests that the reactants are favored.

Understanding \( K_p \) is key to foreseeing how changes will affect the system.

Gibbs Free Energy

Gibbs Free Energy, \( \Delta G \), is the measure of available energy a system has to do work at constant temperature and pressure. It's a pivotal concept in determining the spontaneity of a reaction. In terms of equilibrium, the change in Gibbs Free Energy, \( \Delta G^\circ \), is essential. The relationship between \( \Delta G^\circ \) and the equilibrium constant \( K \) is described by the equation:\[ \Delta G^\circ = -RT \ln K \], where \( R \) is the gas constant and \( T \) is the temperature in Kelvin.

• If \( \Delta G^\circ < 0 \), the reaction is thermodynamically favorable (spontaneous) in the forward direction.
• If \( \Delta G^\circ > 0 \), the reaction is not favorable in the forward direction.

This concept helps link thermodynamics and equilibrium, guiding us on reaction spontaneity.

Le Chatelier's Principle

Le Chatelier's Principle describes how a reaction at equilibrium responds to external changes, like concentration, pressure, or temperature. When a system experiences a disturbance, it shifts its position to counteract the change and re-establish equilibrium. For instance, if in our reaction, more \( \text{CO}_2 \) or \( \text{H}_2 \) is added, the system will shift to the right to produce more \( \text{CO} \) and \( \text{H}_2\text{O} \), counteracting the increase. Conversely, removing product gases like \( \text{CO} \) or increasing pressure will also favor the forward reaction. This principle helps predict how the magnitude of external changes affects the direction and extent of a chemical reaction's shift, maintaining the balance of the system.

Partial Pressure

Partial pressure is the pressure exerted by a single gas within a mixture, comparable to what the gas would exert if it were alone in the container. Each gas in a mixture contributes to the total pressure of the system. In the reaction \( \text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g) \), knowing the partial pressures helps calculate the reaction quotient \( Q_p \) and equilibrium constant \( K_p \). Calculating partial pressure involves:

• The mole fraction of the gas in the mixture.
• The total pressure of the system.

Using partial pressures, we refill the intricate details for gases interacting under given conditions, impacting equilibrium calculations.

Ideal Gas Law

The Ideal Gas Law is a cornerstone formula in chemistry, describing the behavior of gases. It relates the pressure, volume, temperature, and number of moles of a gas with the equation:\[ PV = nRT \], where

• \( P \) is the pressure of the gas,
• \( V \) is the volume,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant (8.314 J/mol·K),
• \( T \) is the temperature in Kelvin.

In the context of chemical equilibrium, it's crucial for determining the partial pressures of gases when analyzing equilibria involving gaseous reactants and products. Understanding this law enables us to interconvert between volume, number of moles, and partial pressures, reflecting how gases will behave under different conditions.