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 \(\mathrm{C}\) 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

The enthalpy change, ΔH° for the reaction of CO with H2 to form methanol is -90.5 kJ/mol. To maximize the equilibrium yield of methanol, a low temperature and high pressure should be used.

Step-by-step solution

Part A: Calculation of Enthalpy Change ΔH°

To calculate the enthalpy change, we can use the data provided in the appendix C for the formation of each compound involved. The equation for calculating ΔH° is: \[\Delta H^{\circ} = \sum_{products}(\Delta H^{\circ}_{formation}) - \sum_{reactants}(\Delta H^{\circ}_{formation})\] Given the data from Appendix C (in kJ/mol): \[\Delta H^{\circ}_{formation}(\mathrm{CH}_{3}\mathrm{OH}) = -201.0\] \[\Delta H^{\circ}_{formation}(\mathrm{CO}) = -110.5\] \[\Delta H^{\circ}_{formation}(\mathrm{H}_{2}) = 0\] Using the equation above, we have: \[\Delta H^{\circ} = (-201.0) - [(-110.5) + 0 * 2]\] \[\Delta H^{\circ} = -90.5 \: \text{kJ/mol}\] The enthalpy change, ΔH° is -90.5 kJ/mol for this reaction.

Part B: Temperature and Equilibrium Yield

According to Le Chatelier's principle, if the reaction is exothermic (ΔH° < 0), increasing the temperature decreases the yield of the product, while decreasing the temperature increases the yield of the product. In this case, ΔH° is negative, which means the reaction is exothermic. To maximize the equilibrium yield of methanol, we should use a low temperature.

Part C: Pressure and Equilibrium Yield

According to Le Chatelier's principle, increasing the pressure of a system at equilibrium favors the side with fewer moles of gas. In this case, the moles of gas in the reaction are: \[\text{Reactants:} \: 1\: (\text{CO}) + 2\:(\text{H}_{2}) = 3 \: \text{moles}\] \[\text{Products:} \: 1\: (\text{CH}_{3}\text{OH}) = 1\:\text{mole}\] To maximize the equilibrium yield of methanol, we should use a high pressure since there are fewer moles of gas on the product side.

Key concepts

Enthalpy Change

In chemical reactions, enthalpy change (\(\Delta H^{\circ} \)) describes the heat absorbed or released. It's crucial to determine whether a reaction is exothermic (releases heat) or endothermic (absorbs heat). Calculating the enthalpy change involves finding the difference between the enthalpy of formation of products and reactants. For the methanol synthesis reaction: \[\Delta H^{\circ} = \sum_{\text{products}}(\Delta H^{\circ}_{\text{formation}}) - \sum_{\text{reactants}}(\Delta H^{\circ}_{\text{formation}})\]In our specific case:

• \(\Delta H^{\circ}_{\text{formation}}(\mathrm{CH}_{3}\mathrm{OH}) = -201.0 \text{ kJ/mol}\)
• \(\Delta H^{\circ}_{\text{formation}}(\mathrm{CO}) = -110.5 \text{ kJ/mol}\)
• \(\Delta H^{\circ}_{\text{formation}}(\mathrm{H}_{2}) = 0 \text{ kJ/mol}\)

Substituting these values into our formula gives us an enthalpy change of:\[\Delta H^{\circ} = (-201.0) - [(-110.5) + 0 \times 2] = -90.5 \text{ kJ/mol}\]A negative value indicates an exothermic reaction, demonstrating that methanol formation releases energy as heat.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental chemical concept that helps predict how changes in conditions affect chemical equilibrium. When a system at equilibrium experiences a change in temperature, pressure, or concentration, the principle states that the system will adjust to counteract that change, attempting to restore equilibrium.For temperature changes:- If a reaction is exothermic (\(\Delta H^{\circ} < 0\)), increasing temperature will favor the endothermic reverse reaction. - Thus, lowering the temperature shifts the equilibrium toward the exothermic forward reaction, increasing product yield.In the context of methanol production, since the reaction is exothermic (\(-90.5 \text{ kJ/mol}\)), we should lower the temperature to favor the formation of methanol. This helps maximize the equilibrium yield by shifting more CO and H2 into methanol.Understanding the impacts of temperature changes allows chemists to optimize conditions for maximum yield, emphasizing the importance of Le Chatelier's Principle in industrial applications.

Equilibrium Yield

Equilibrium yield refers to the amount of product formed when a reaction reaches equilibrium. It's vital in industrial settings to maximize the output of desirable products. According to Le Chatelier's Principle, one of the ways to enhance yield is adjusting pressure, especially in reactions involving gases.In our methanol synthesis:- We have 3 moles of gas on the reactant side (1 CO and 2 H\(_{2}\))- We have 1 mole of gas on the product side (CH\(_{3}\)OH)Increasing pressure shifts the equilibrium to the side with fewer gas moles. Therefore, in this reaction, higher pressure favors the formation of methanol, reducing reactant moles by efficiently pushing more towards products.These principles are crucial because they allow industries to finely tune reaction conditions, achieving a higher equilibrium yield and improving overall productivity. Empowered with this knowledge, chemists can design optimum conditions to ensure maximum efficiency.