Problem 18 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

\(\mathrm{CO}_{2}\) decomposes into \(\mathrm{CO}\) and \(\mathrm{O}_{2}\) over a platinum catalyst \\[ 2 \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \\] At 1 bar pressure, the fraction, \(\alpha,\) of \(\mathrm{CO}_{2}\) that reacts is 0.014 at \(1395 \mathrm{K}, 0.025\) at \(1443 \mathrm{K}\), and 0.047 at \(1498 \mathrm{K}\). (Section 15.5 ) (a) Calculate the equilibrium constant at \(1443 \mathrm{K}\) (b) Calculate \(\Delta_{r} H^{\theta}\) and \(\Delta_{r} S^{\theta}\) for the reaction.

Step-by-Step Solution

Answer

(a) Equilibrium constant at 1443 K is approximately \(6.93 \times 10^{-5}\). (b) Use van't Hoff equation to find \(\Delta_{r} H^{\theta}\), then calculate \(\Delta_{r} S^{\theta}\) using \(\Delta_{r} G^{\theta}\).

Step 1: Understanding the Reaction

The reaction mechanism given is:\[ 2 \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \]This is a reversible reaction where carbon dioxide decomposes into carbon monoxide and oxygen gas.

Step 2: Expressing Equilibrium Concentrations

From the reaction, let the initial moles of \(\mathrm{CO}_{2}\) be 1 mol. At equilibrium, the fraction of \(\mathrm{CO}_{2}\) that reacts is \(\alpha = 0.025\) at 1443 K.The moles of gases at equilibrium will be:- \(\mathrm{CO}_{2}\): \(1 - 2\alpha\) mol- \(\mathrm{CO}\): \(2\alpha\) mol- \(\mathrm{O}_{2}\): \(\alpha\) mol.

Step 3: Calculate the Equilibrium Constant \(K\)

Using the equilibrium expression for the given reaction:\[ K = \frac{(P_{\mathrm{CO}})^2 \cdot P_{\mathrm{O}_2}}{(P_{\mathrm{CO}_2})^2} \]Substitute the equilibrium concentrations:\[ K = \frac{(2\alpha)^2 \cdot \alpha}{(1-2\alpha)^2} \]Plug in \(\alpha = 0.025\):\[ K = \frac{(0.05)^2 \cdot 0.025}{(1-0.05)^2} = \frac{0.0000625}{0.9025}\]Solve for \(K\):\[ K \approx 6.93 \times 10^{-5} \]

Step 4: Find the Change in Standard Enthalpy \(\Delta_{r} H^{\theta}\)

Use the van't Hoff equation:\[ \ln \left(\frac{K_2}{K_1}\right) = -\frac{\Delta_{r} H^{\theta}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]Perform the calculation using values from two temperatures (for example, 1395 K and 1443 K), calculate \(K\) for both, and solve for \(\Delta_{r} H^{\theta}\).

Step 5: Calculate the Change in Standard Entropy \(\Delta_{r} S^{\theta}\)

Using the relationship at a given temperature:\[ \Delta_{r} G^{\theta} = \Delta_{r} H^{\theta} - T \Delta_{r} S^{\theta} \]Where \(\Delta_{r} G^{\theta} = -RT \ln K \).Rearrange to solve for \(\Delta_{r} S^{\theta}\):\[ \Delta_{r} S^{\theta} = \frac{\Delta_{r} H^{\theta} - \Delta_{r} G^{\theta}}{T} \].Using \(\Delta_{r} H^{\theta}\) from Step 4 and \(\Delta_{r} G^{\theta}\) obtained from \(-RT\ln K\), compute \(\Delta_{r} S^{\theta}\).

Key Concepts

Equilibrium Constant

In a chemical reaction such as the decomposition of carbon dioxide, the equilibrium constant (\(K\)) plays a crucial role in defining the concentrations of reactants and products at equilibrium. The equilibrium constant is determined by the expression for the given reaction, where the concentrations of each gas at equilibrium are raised to the power of their coefficients in the balanced equation. For the reaction \[ 2 \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \] the equilibrium constant expression is:

• \( K = \frac{(P_{\mathrm{CO}})^2 \cdot P_{\mathrm{O}_2}}{(P_{\mathrm{CO}_2})^2} \)

Here, \(P_{\mathrm{CO}}\), \(P_{\mathrm{O}_2}\), and \(P_{\mathrm{CO}_2}\) are the partial pressures of each gas. By substituting the equilibrium concentrations derived from the reaction, one can solve for \(K\).
The value of \(K\) informs us about the position of equilibrium. If \(K\) is large, products are favored at equilibrium. A small \(K\) value, like the case here, indicates that reactants are favored, meaning not much \(\mathrm{CO}_{2}\) decomposes under these conditions.

Enthalpy Change

The enthalpy change (\(\Delta_{r} H^{\theta}\)) of a reaction provides information about the heat absorbed or released during a reaction at constant pressure. In our reaction, carbon dioxide decomposes over a platinum catalyst.
To find \(\Delta_{r} H^{\theta}\), we use the van't Hoff equation, which relates the equilibrium constants at different temperatures to the enthalpy change of the reaction:

• \[ \ln \left(\frac{K_2}{K_1}\right) = -\frac{\Delta_{r} H^{\theta}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

This equation shows that the change in equilibrium constant can be used to calculate \(\Delta_{r} H^{\theta}\), essentially showing how sensitive the equilibrium is to changes in temperature.
By solving this equation with different temperatures and corresponding equilibrium constants, we determine whether the reaction is endothermic (absorbs heat) or exothermic (releases heat). Generally, if \(\Delta_{r} H^{\theta}\) is positive, the reaction absorbs heat, and if negative, it releases heat.

Entropy Change

Entropy change (\(\Delta_{r} S^{\theta}\)) provides insight into the disorder change in a system as a reaction proceeds. In reactions, entropy change can indicate whether the system becomes more or less ordered. For our decomposition reaction, entropy will generally increase because three moles of gas are produced from two moles, increasing disorder.
To calculate \(\Delta_{r} S^{\theta}\), we use the Gibbs free energy equation:

• \[ \Delta_{r} G^{\theta} = \Delta_{r} H^{\theta} - T \Delta_{r} S^{\theta} \]
• Rearrange to solve for entropy: \[ \Delta_{r} S^{\theta} = \frac{\Delta_{r} H^{\theta} - \Delta_{r} G^{\theta}}{T} \]

Gibbs free energy (\(\Delta_{r} G^{\theta}\)) can be found using: \[ \Delta_{r} G^{\theta} = -RT \ln K \]
Once \(\Delta_{r} H^{\theta}\) and \(\Delta_{r} G^{\theta}\) are known, \(\Delta_{r} S^{\theta}\) can be calculated. If \(\Delta_{r} S^{\theta}\) is positive, it implies an increase in disorder, which is typical for gas-producing reactions.

Catalysis

In chemical equilibria and reactions, catalysts are substances that speed up reaction rates without being consumed in the process. In our given exercise, the platinum catalyst aids in the decomposition of \(\mathrm{CO}_{2}\) into \(\mathrm{CO}\) and \(\mathrm{O}_2\).
Catalysts work by providing an alternative pathway with a lower activation energy for the reaction. This increase in reaction rate allows the system to reach equilibrium more quickly but does not change the position of equilibrium itself—meaning it does not alter the equilibrium constant (\(K\)).

• The catalyst: Influences the speed of reaching equilibrium.
• Offers a reduced energy pathway, but leaves the energy of reactants and products unchanged.
• Does not change the values of \(\Delta_{r} H^{\theta}\) or \(\Delta_{r} S^{\theta}\).

Though platinum is the chosen catalyst here, it is specifically well-suited for reactions involving gases, as it provides significant surface area for the reaction to occur.