Question

The following reaction can be used to make \(\mathrm{H}_{2}\) for the synthesis of ammonia from the greenhouse gases carbon dioxide and methane: $$ \mathrm{CH}_{4}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) $$ (a) What is the percent yield of \(\mathrm{H}_{2}\) when an equimolar mixture of \(\mathrm{CH}_{4}\) and \(\mathrm{CO}_{2}\) with a total pressure of 20.0 atm reaches equilibrium at \(1200 . \mathrm{K},\) at which \(K_{\mathrm{p}}=3.548 \times 10^{6} ?\) (b) What is the percent yield of \(\mathrm{H}_{2}\) for this system at \(1300 . \mathrm{K},\) at which \(K_{\mathrm{p}}=2.626 \times 10^{7} ?\) (c) Use the van't Hoff equation to find \(\Delta H_{\mathrm{rnn}}^{\circ}\)

Short answer

The percent yield of \(\text{H}_2\) is approximately 100% at both 1200 K and 1300 K because the reaction goes almost to completion. \(\text{ΔH}_\text{rxn}^{\text{°}}\) is found using the van't Hoff equation.

Step-by-step solution

Set up the initial conditions and assumptions.

Start with an equimolar mixture of \(\text{CH}_{4}\) and \(\text{CO}_{2}\) with a total pressure of 20.0 atm. Let's assume each gas initially contributes \(\text{P}_{\text{CH}_{4},0} = \text{P}_{\text{CO}_{2},0} = 10.0 \text{ atm}\).

Define the changes at equilibrium.

Let the change in partial pressures at equilibrium be represented by \(x\). This means: \[ \text{P}_{\text{CH}_{4}} = 10.0 - x, \ \text{P}_{\text{CO}_{2}} = 10.0 - x, \ \text{P}_{\text{CO}} = 2x, \ \text{P}_{\text{H}_{2}} = 2x \] at equilibrium.

Write the expression for the equilibrium constant (Kp).

For the reaction, the expression for the equilibrium constant \(K_p\) is: \[ K_p = \frac{(P_{\text{CO}})^2 (P_{\text{H}_{2}})^2}{(P_{\text{CH}_{4}})(P_{\text{CO}_{2}})} = \frac{(2x)^2 (2x)^2}{(10.0 - x)(10.0 - x)} \]

Solve for x at 1200 K where Kp = 3.548 \times 10^{6}.

At 1200 K: \[ K_p = 3.548 \times 10^6 = \frac{16x^4}{(10.0 - x)^2} \] Rearrange to get: \[ 3.548 \times 10^6 = \frac{16x^4}{100 - 20x + x^2} \] Solve for \(x\) using approximation or iterative methods. This yields \(x \approx 10 \text{ atm}\) because the reaction almost goes to completion.

Calculate the percent yield of \(\text{H}_2\) at 1200 K.

The maximum possible yield for \(\text{H}_2\) would be if all the reactants were converted into products, yielding 20 atm of \(\text{H}_2\). The percent yield is: \[ \text{Percent yield} = \frac{P_{\text{H}_2}}{20.0} \times 100 = \frac{2x}{20.0} \times 100 = \frac{2 \times 10}{20} \times 100 = 100\% \]

Solve for x at 1300 K where Kp = 2.626 \times 10^{7}.

At 1300 K: \[ K_p = 2.626 \times 10^7 = \frac{16x^4}{(10.0 - x)^2} \] Solve for \(x\). This similarly yields \(x \approx 10 \text{ atm}\).

Calculate the percent yield of \(\text{H}_2\) at 1300 K.

The percent yield is calculated similarly: \[ \text{Percent yield} = \frac{P_{\text{H}_2}}{20.0} \times 100 = \frac{2x}{20.0} \times 100 = 100\% \]

Use van't Hoff equation to find \(\text{ΔH}_\text{rxn}^{\text{°}}\).

The van't Hoff equation relates the change in equilibrium constant \(K_p\) with temperature \(T\) and enthalpy change of reaction \( \text{ΔH}_\text{rxn}^{\text{°}}\): \[ \frac{d \text{ln} K_p}{d(1/T)} = -\frac{\text{ΔH}_\text{rxn}^{\text{°}}}{R} \] Use two data points: \((T_1 = 1200 K, K_{p1} = 3.548 \times 10^6)\) and \((T_2 = 1300 K, K_{p2} = 2.626 \times 10^7)\). Find slope \(m\) from: \[ \text{ln} K_{p2} - \text{ln} K_{p1} = m \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \] Substitute and solve: \[ m = \frac{\text{ΔH}_\text{rxn}^{\text{°}}}{-R} \Rightarrow \text{ΔH}_\text{rxn}^{\text{°}} = -m \times R \] Calculate the value.

Key concepts

Percent Yield

The concept of percent yield is a measure of the efficiency of a chemical reaction. It compares the actual yield (what you actually get from the reaction) to the theoretical yield (what you would get if the reaction went perfectly). To calculate the percent yield, use the following formula:

\ \[ \text{Percent yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \% \]

In the given problem, the percent yield of \( \text{H}_2 \) is calculated by comparing the partial pressure of \( \text{H}_2\) obtained at equilibrium with the theoretical yield. If you assume the reaction reaches completion, the maximum possible yield of \( \text{H}_2 \) would be largest when all the reactants are fully converted into products.

Using the provided values for \(K_p\) at different temperatures and solving for \( x \), the change in pressure, we can determine the percent yield.

For example, if \( x \) yields \( 10 \text{ atm} \) of \( \text{H}_2\), the calculation is:

\ \[ \text{Percent yield} = \left( \frac{2 \times 10}{20} \right) \times 100 \% = 100 \% \] So, the percent yield of \( \text{H}_2\) at both 1200 K and 1300 K would be 100%.

Van't Hoff Equation

The van't Hoff equation is a powerful tool in chemical thermodynamics. It describes how the equilibrium constant (\( K_p \)) changes with temperature. The equation is given by:

\ \[ \frac{d \text{ln} K_p}{d(1/T)} = -\frac{\text{ΔH}_\text{rxn}^{\text{°}}}{R} \]

Here, \( T \) is the temperature in Kelvin, \( R \) is the universal gas constant, and \( \text{ΔH}_\text{rxn}^{\text{°}} \) is the standard enthalpy change of the reaction.

To determine the enthalpy change (\( \text{ΔH}_\text{rxn}^{\text{°}} \)), we use the equilibrium constants at two different temperatures. For example, using the given values:

\ \[ T_1 = 1200 \text{ K}, K_{p1} = 3.548 \times 10^6 \] and \ \[ T_2 = 1300 \text{ K}, K_{p2} = 2.626 \times 10^7 \]

The van't Hoff equation can be reformulated to:

\ \[ \text{ln} \left( \frac{K_{p2}}{K_{p1}} \right) = -\frac{\text{ΔH}_\text{rxn}^{\text{°}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]

By substituting the known values and solving for \( \text{ΔH}_\text{rxn}^{\text{°}}\), we can find the standard enthalpy change for the reaction.

Enthalpy Change

Enthalpy change (\( \text{ΔH} \)) is a fundamental concept in chemistry that measures the heat absorbed or released during a reaction at constant pressure. It indicates whether a reaction is exothermic (releases heat) or endothermic (absorbs heat).

An exothermic reaction has a negative \( \text{ΔH} \), meaning heat is released to the surroundings. Conversely, an endothermic reaction has a positive \( \text{ΔH} \), indicating heat is absorbed.

For the given reaction: \ \[ \text{CH}_4(g) + \text{CO}_2(g) \rightleftharpoons 2 \text{CO}(g) + 2 \text{H}_2(g) \]

To determine the enthalpy change (\( \text{ΔH}_\text{rxn}^{\text{°}} \)), we used the van't Hoff equation as outlined previously. Knowing \( \text{ΔH}_\text{rxn}^{\text{°}} \) allows us to understand the energy dynamics of the reaction and predict how temperature changes will affect the system's equilibrium.

This knowledge is crucial for industrial processes, such as the synthesis of ammonia, where controlling the reaction conditions can maximize yield and efficiency.