Question

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the reaction of \(\mathrm{CO}\) with \(\mathrm{H}_{2} :\) $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ (a) Use thermochemical data in Appendix C to calculate \(\Delta H^{\circ}\) for this reaction. (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature? (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?

Short answer

(a) The standard enthalpy change for the reaction, $\Delta H^{\circ}$, is calculated as: \[ \Delta H^{\circ} = \Sigma n \Delta H_{f}^{\circ}(\text{products}) - \Sigma m \Delta H_{f}^{\circ}(\text{reactants}) \] (b) Based on the calculated ∆H°, if the reaction is exothermic (∆H° < 0), low temperature would maximize the equilibrium yield of methanol; if the reaction is endothermic (∆H° > 0), high temperature would maximize the equilibrium yield of methanol. (c) High pressure conditions would maximize the equilibrium yield of methanol, since it shifts the equilibrium towards the side with fewer moles of gas, favoring the production of methanol.

Step-by-step solution

a) Calculate the standard enthalpy change (∆H°) for the reaction

To calculate the standard enthalpy change for the reaction, we need to look up the standard enthalpies of formation (∆Hf°) for CO, H2, and CH3OH in Appendix C. Next, use the following equation to calculate ∆H°: \[ \Delta H^{\circ} = \Sigma n \Delta H_{f}^{\circ}(\text{products}) - \Sigma m \Delta H_{f}^{\circ}(\text{reactants}) \]

b) Determine the temperature conditions for maximum methanol yield

According to Le Chatelier's principle, if the reaction is exothermic (∆H° < 0), high temperature would shift the equilibrium to the left, favoring reactants. Conversely, if the reaction is endothermic (∆H° > 0), high temperature would shift the equilibrium to the right, favoring products. Based on the calculated ∆H°, we can determine whether high or low temperature conditions would maximize the equilibrium yield of methanol.

c) Determine the pressure conditions for maximum methanol yield

According to Le Chatelier's principle, increasing the pressure will shift the equilibrium towards the side with fewer moles of gas. In our case, there are 3 moles of gas on the reactant side (1 mol CO + 2 mol H2) and 1 mole of gas on the product side (CH3OH). Increasing pressure will shift the equilibrium towards the side with fewer moles of gas, i.e., to the right, which will favor the production of methanol. So, high pressure conditions would maximize the equilibrium yield of methanol.

Key concepts

Le Chatelier's Principle

Le Chatelier's Principle helps us understand how a chemical reaction at equilibrium responds to changes in conditions. The principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract that change.

• For an exothermic reaction (\(\Delta H^\circ < 0\)), increasing the temperature shifts the equilibrium toward the reactants.
• For an endothermic reaction, increasing the temperature shifts the equilibrium toward the products.

This principle is helpful for predicting the behavior of reactions when conditions like temperature and pressure change.

Equilibrium Yield

The equilibrium yield refers to the amount of product formed when the reaction reaches equilibrium. To maximize the yield of a particular product, it's crucial to apply Le Chatelier's principle.

For methanol production, the equilibrium yield is affected by both temperature and pressure:

• Temperature: If the reaction releases heat (exothermic), lowering the temperature would favor the formation of methanol.
• Pressure: Increasing the pressure can shift the equilibrium toward the product side since there are fewer moles of gas.

By optimizing these conditions, we can increase the production efficiency.

Pressure Effects

Pressure changes can significantly influence the position of equilibrium in gas-phase reactions. For reactions involving gases, increasing the pressure will favor the side with fewer moles of gas molecules.

In the methanol production reaction:
\(\text{CO}(g) + 2\,\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g)\) has 3 moles of reactants and 1 mole of product.

• Under high pressure, the equilibrium shifts to the right, favoring methanol formation.

This is why high-pressure conditions are often used in industrial reactions to enhance yield.

Thermochemistry Calculations

Thermochemistry calculations involve determining the heat exchange in chemical reactions. The standard enthalpy change (\(\Delta H^\circ\)) is a primary focus, representing the heat absorbed or released at constant pressure.

To calculate \(\Delta H^\circ\) for a reaction:

• Look up the standard enthalpies of formation \((\Delta H_f^\circ)\) for all reactants and products.
• Use the formula: \(\Delta H^\circ = \Sigma n \Delta H_f^\circ(\text{products}) - \Sigma m \Delta H_f^\circ(\text{reactants})\).

This calculation helps us understand the thermal in nature of reactions, enabling better control of reaction conditions.