Problem 21 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

One stage in the manufacture of methanol from methane involves the conversion of synthesis gas (a mixture of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) to methanol. The conversion is carried out over a catalyst at a temperature of around \(500 \mathrm{K}\) and a pressure of 100 atm. (Section \(1.9)\) $$\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g}) \Delta H=-90.7 \mathrm{kJmol}^{-1}$$ (a) Write an expression for \(K_{p}\) for the reaction. (b) At \(500 \mathrm{K}\) and 100 atm pressure, an equilibrium mixture contains \(42 \% \mathrm{CH}_{3} \mathrm{OH}\) and \(48 \% \mathrm{CO}\). Calculate a value for \(K_{p}\) at this temperature. (c) Use Le Chatelier's principle to predict what would happen to the percentage of methanol in the mixture if: (i) the temperature increases; (ii) the pressure increases; (iii) hydrogen is added at constant temperature and pressure.

Step-by-Step Solution

Answer

(a) \(K_p = \frac{P_{\mathrm{CH}_3\mathrm{OH}}}{P_{\mathrm{CO}} \cdot (P_{\mathrm{H}_2})^2}\), (b) \(K_p \approx 0.00875\), (c) (i) Decrease, (ii) Increase, (iii) Increase methanol percentage.

Step 1: Expression for Kp

The equilibrium constant expressed in terms of pressure, \(K_{p}\), for the given reaction \(\mathrm{CO} + 2\mathrm{H}_2 \rightleftharpoons \mathrm{CH}_3\mathrm{OH}\) is derived from the partial pressures of the reactants and products at equilibrium.For the reaction, \(K_{p}\) is given by:\[ K_{p} = \frac{P_{\mathrm{CH}_3\mathrm{OH}}}{P_{\mathrm{CO}} \cdot (P_{\mathrm{H}_2})^2} \]

Step 2: Calculate Partial Pressures

The total pressure of the system is 100 atm. Given that the equilibrium mixture is 42% \(\mathrm{CH}_{3}\mathrm{OH}\), 48% \(\mathrm{CO}\), and the rest is \(\mathrm{H}_{2}\), the partial pressures can be calculated:- \(P_{\mathrm{CH}_3\mathrm{OH}} = 0.42 \times 100 = 42 \) atm- \(P_{\mathrm{CO}} = 0.48 \times 100 = 48 \) atm- The remaining percentage for \(\mathrm{H}_2\) is \(100\% - 42\% - 48\% = 10\%\), so \(P_{\mathrm{H}_2} = 0.10 \times 100 = 10 \) atm

Step 3: Substitute Values into Kp Expression

Substitute the calculated partial pressures into the \(K_{p}\) expression:\[ K_{p} = \frac{42}{48 \times 10^2} = \frac{42}{4800} \approx 0.00875 \]

Step 4: Le Chatelier's Principle - Temperature Influence

According to Le Chatelier's principle, if the temperature increases, the equilibrium will shift to the endothermic side to absorb heat. Given \(\Delta H = -90.7 \mathrm{kJ/mol}\), the reaction is exothermic, so increasing the temperature will decrease the percentage of \(\mathrm{CH}_{3}\mathrm{OH}\) in the mixture.

Step 5: Le Chatelier's Principle - Pressure Influence

An increase in pressure will make the equilibrium shift towards the side with fewer moles of gas. The reactant side has 3 moles (1 \(\mathrm{CO}\), 2 \(\mathrm{H}_2\)) whereas the product side \(\mathrm{CH}_{3}\mathrm{OH}\) has 1 mole. Thus, increasing the pressure favors the production of \(\mathrm{CH}_{3}\mathrm{OH}\), increasing its percentage.

Step 6: Le Chatelier's Principle - Adding Hydrogen

Adding hydrogen at constant temperature and pressure will favor the forward reaction since it shifts the equilibrium to decrease the added species (\(\mathrm{H}_2\)). This will increase the percentage of \(\mathrm{CH}_{3}\mathrm{OH}\) in the mixture.

Key Concepts

Le Chatelier's Principle

Le Chatelier's Principle helps us predict how a chemical equilibrium will respond to changes in conditions. It's like knowing how a scale will tip if we add or remove weight from either side.

When we talk about temperature, pressure or concentration changes, this principle is our guide. For example, if the temperature of an exothermic reaction, like our methanol synthesis, increases, the equilibrium will shift to absorb the added heat by favoring the endothermic direction.

• **Temperature Increase**: An increase in temperature will cause the system to favor the reaction that absorbs heat. Since methanol synthesis is exothermic, producing heat, raising the temperature will shift the equilibrium to the left, thus reducing methanol production.


• **Pressure Increase**: The system wants to minimize changes. If we increase pressure, the equilibrium shifts towards the side with fewer gas moles. The methanol side has fewer moles, so increasing pressure favors its production.


• **Adding a Reactant**: If hydrogen is added, the system will work to reduce its concentration by increasing the production of methanol, thereby shifting equilibrium toward the products.

Equilibrium Constant (Kp)

The equilibrium constant, denoted as **Kp**, is related to the partial pressures of gases involved in a chemical reaction at equilibrium. It gives us a snapshot of the reaction's dynamics at equilibrium.

For the reaction of carbon monoxide and hydrogen to form methanol, the expression for Kp is: \[ K_{p} = \frac{P_{\mathrm{CH}_3\mathrm{OH}}}{P_{\mathrm{CO}} \cdot (P_{\mathrm{H}_2})^2} \]

This formula shows how the pressures of the reactants and products interrelate at equilibrium. A small Kp value, like 0.00875 from our example, indicates that the reactants are favored at equilibrium. In contrast, a larger Kp would suggest that the products are favored.

Understanding Kp is crucial for adjusting conditions (temperature, pressure) effectively to optimize product yield, like increasing methanol in our process.

Catalysis

Catalysis involves the use of a catalyst to accelerate a reaction without being consumed by it. In our methanol synthesis example, a catalyst helps reach equilibrium faster but does not shift the equilibrium position.

A catalyst works by opening a new reaction pathway with a lower activation energy. This is especially useful at industrial scales, where reaction time is money. Although catalysts do not appear in the Kp expression or alter the equilibrium concentrations, they are essential tools for optimizing the practical application of reactions.

Different catalysts, such as metal oxides, might be specifically used to enhance the conversion process from synthesis gas to methanol at high pressures and temperatures like the ones in this reaction.

Partial Pressure

Partial pressure is the pressure exerted by a single gas component in a mixture of gases. For chemical equilibria involving gases, understanding partial pressures helps us calculate Kp and predict reaction behavior.

In our methanol production example, each gas contributes to the total pressure of 100 atm. Here's how:

• **Methanol (CH₃OH)**: 42% corresponds to 42 atm of pressure.


• **Carbon Monoxide (CO)**: 48% corresponds to 48 atm.


• **Hydrogen (H₂)**: The remaining 10% gives 10 atm.

The relation between these pressures under fixed conditions tells us how the equilibrium might shift with changes like pressure increases. Accurately identifying these partial pressures is vital for calculating the equilibrium constant and foreseeing the effects of altering reaction conditions.