Problem 31 - Chemistry An Atoms First Approach

Question

At \(327^{\circ} \mathrm{C},\) the equilibrium concentrations are \(\left[\mathrm{CH}_{3} \mathrm{OH}\right]=\) \(0.15 M,[\mathrm{CO}]=0.24 M,\) and \(\left[\mathrm{H}_{2}\right]=1.1 M\) for the reaction $$\mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)$$.Calculate \(K_{\mathrm{p}}\) at this temperature.

Step-by-Step Solution

Answer

The equilibrium constant \(K_p\) for the reaction \(\mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)\) at 327°C is approximately 56207.13.

Step 1: Write down the given information

We are given the following equilibrium concentrations: - [CH3OH] = 0.15 M - [CO] = 0.24 M - [H2] = 1.1 M We also know the temperature, which is 327°C, but we should convert it to Kelvin (K) for our calculations: Temperature = 327 + 273.15 = 600.15 K

Step 2: Calculate the partial pressures using the Ideal Gas Law

To find the partial pressures, we can use the Ideal Gas Law given by: PV = nRT Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 Latm/molK), and T is the temperature in Kelvin. Also, note that for our equilibrium constant, it doesn't matter what the volume of the container is because it is constant for all the species involved so it would cancel out. Calculating partial pressures using the Ideal Gas Law and replacing the concentrations with n/V, we get: P(CH3OH) = [CH3OH]RT P(CO) = [CO]RT P(H2) = [H2]RT Plug in the given concentrations and the temperature: P(CH3OH) = (0.15 M)(0.0821 Latm/molK)(600.15 K) = 7.37 atm P(CO) = (0.24 M)(0.0821 Latm/molK)(600.15 K) = 11.81 atm P(H2) = (1.1 M)(0.0821 Latm/molK)(600.15 K) = 54.05 atm

Step 3: Use the equilibrium expression to find Kp

The expression for Kp for the given reaction is: Kp = (P(CO) * P(H2)^2) / P(CH3OH) Plug in the calculated partial pressures: Kp = (11.81 atm * (54.05 atm)^2) / 7.37 atm = 56207.13 So, the equilibrium constant Kp for the reaction CH3OH(g) ⇌ CO(g) + 2 H2(g) at 327°C is approximately 56207.13.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. This equation is expressed as \( PV = nRT \). Here, \( P \) stands for pressure, \( V \) is volume, \( n \) represents the number of moles, \( R \) is the gas constant, and \( T \) is the temperature measured in Kelvin.

When working with gases in a chemical equilibrium problem, like the one given, we often use the Ideal Gas Law to calculate partial pressures from molar concentrations. This is particularly useful because equilibrium constants in gas reactions, often represented as \( K_p \), are expressed in terms of pressure. The transformation assumes that concentration (moles per volume) can be related to pressure using this law.

To make these calculations, you must first convert temperature to Kelvin, as Kelvin is the temperature unit required in this equation. With temperature set and given concentrations, you can compute the pressures of gases involved as follows: \( P = [ ext{Concentration}] \times R \times T \). This approach allows us to transition seamlessly between concentration and pressure.

Partial Pressure Calculation

In chemistry, calculating partial pressures is a crucial step in understanding gas reactions at equilibrium. Partial pressure refers to the pressure that a specific gas in a mixture would exert if it were alone in the container.

Using the Ideal Gas Law, we can transform molar concentrations into partial pressures. This is done by multiplying the concentration of a gas by the product of the universal gas constant \( R \) and the temperature in Kelvin. As shown above, the formula is \( P = [ ext{Concentration}] \times R \times T \).

• For \( \text{CH}_3\text{OH} \): \( P = 0.15 \times 0.0821 \times 600.15 = 7.37 \text{ atm} \)
• For \( \text{CO} \): \( P = 0.24 \times 0.0821 \times 600.15 = 11.81 \text{ atm} \)
• For \( \text{H}_2 \): \( P = 1.1 \times 0.0821 \times 600.15 = 54.05 \text{ atm} \)

These values are crucial for calculating the equilibrium constant \( K_p \), which involves the use of these partial pressures.

Chemical Equilibrium Calculation

Chemical equilibrium refers to a state in a chemical reaction where the rates of the forward and reverse reactions are equal, resulting in stable concentrations of reactants and products. The equilibrium constant, \( K_p \), is a value that reflects the ratio of the pressures of products to reactants at equilibrium for gaseous reactions.

For reactions involving gases, \( K_p \) is defined using partial pressures rather than concentrations. For the reaction \( \text{CH}_3\text{OH} \rightleftharpoons \text{CO} + 2 \text{H}_2 \), the expression for \( K_p \) is given by:\[K_p = \frac{P(\text{CO}) \times P(\text{H}_2)^2}{P(\text{CH}_3\text{OH})}\]

Substitute the calculated partial pressures:\[K_p = \frac{11.81 \times (54.05)^2}{7.37} = 56207.13\]

This calculated \( K_p \) helps predict the direction of the reaction and the extent to which products and reactants coexist at equilibrium. A high \( K_p \) value, such as 56207.13, indicates a greater concentration of products compared to reactants, signifying that the reaction strongly favors the formation of products at this specific temperature.