Question

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the reaction of \(\mathrm{CO}\) with \(\mathrm{H}_{2}\) : $$ \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ (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 enthalpy change for the methanol formation reaction is \(-90.2 \, kJ/mol\). (b) To maximize the equilibrium yield of methanol, a low temperature should be used. (c) To maximize the equilibrium yield of methanol, a high pressure should be used.

Step-by-step solution

Calculate the Enthalpy Change ΔH°

To calculate the enthalpy change for the reaction, use the thermochemical data from Appendix C. The standard enthalpy change can be calculated using the formula: \[ΔH^{\circ} = Σn_pΔH^{\circ}_p - Σn_rΔH^{\circ}_r\] where \(ΔH^{\circ}_p\) and \(ΔH^{\circ}_r\) are the standard enthalpies of formation of the products and reactants, and \(n_p\) and \(n_r\) are the stoichiometric coefficients of the products and reactants.

Thermochemical data lookup

Look up the standard enthalpies of formation for CO(g), H2(g), and CH3OH(g) in Appendix C: - CO(g): \(-110.5 \, kJ/mol\) - H2(g): \(0 \, kJ/mol\) (elements in their standard states have zero enthalpy of formation) - CH3OH(g): \(-200.7 \, kJ/mol\)

Calculate ΔH° for the reaction

Substitute the values found in Step 2 into the formula in Step 1: \[ ΔH^{\circ} = (1)(-200.7 \, kJ/mol) - [(1)(-110.5 \, kJ/mol) + (2)(0 \, kJ/mol)] \] \[ ΔH^{\circ} = -200.7 + 110.5 = -90.2 \, kJ/mol \] The enthalpy change for the formation of methanol is \(-90.2 \, kJ/mol\).

Analyze the effect of temperature on the equilibrium yield

Since the formation of methanol is exothermic (\(ΔH^{\circ} < 0\)), according to Le Chatelier's principle, increasing the temperature will shift the equilibrium to the left (favoring the reactants). Therefore, to maximize the equilibrium yield of methanol, a low temperature should be used.

Analyze the effect of pressure on the equilibrium yield

According to Le Chatelier's principle, increasing the pressure will shift the equilibrium to the side with fewer moles of gas. In the given reaction, there are 3 moles of gas on the reactant side (1 mole of CO plus 2 moles of H2) and only 1 mole of gas on the product side (methanol). Therefore, to maximize the equilibrium yield of methanol, a high pressure should be used. In conclusion, (a) The enthalpy change for the methanol formation reaction is \(-90.2 \, kJ/mol\). (b) To maximize the equilibrium yield of methanol, a low temperature should be used. (c) To maximize the equilibrium yield of methanol, a high pressure should be used.

Key concepts

Thermochemical Data

Thermochemical data, such as standard enthalpies of formation, are critical for calculating the heat changes that occur in chemical reactions. These values are measured under standard conditions, which include a constant pressure of 1 bar and a temperature of 298.15 K (25°C). The standard enthalpy of formation,e>e>e>e>e>e>(e>e>e>e>e>e>e>), is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. For instance, the enthalpy of formation of carbon monoxide (CO(g)) and methanol (CH3OH(g)) are essential to determine the overall heat change for the reaction. It’s important to remember that the reference form of each element, such as H2(g) for hydrogen, has an enthalpy of formation of zero.To solve problems involving enthalpy change, we locate these values in tables like Appendix C. These numbers form the backbone of our calculations, using the formula to find the enthalpy change for a reaction. Knowing how to interpret and apply thermochemical data is an essential skill in chemistry as it allows us to predict the energy requirements or releases in chemical processes.

Le Chatelier's Principle

Le Chatelier's principle helps us understand how a reaction in equilibrium responds to changes in conditions such as temperature, pressure, and concentration. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the change. This behavior allows chemists to manipulate the conditions to favor the production of desired products.For example, when a reaction is exothermic (releases heat), increasing the temperature will cause the equilibrium to shift towards the reactants, decreasing the yield of products. Conversely, reducing the temperature favors the product side and increases the yield in exothermic reactions. The principle applies to pressure changes as well; increasing the pressure on a system where gases are involved will favor the side with fewer moles of gas, as seen in the synthesis of methanol. Le Chatelier's principle is a powerful tool when determining the optimal conditions for industrial chemical processes, ensuring that they are both efficient and economically viable.

Equilibrium Yield

Equilibrium yield refers to the amount of product formed when a chemical reaction reaches a state of dynamic equilibrium. At this point, the forward and reverse reactions occur at the same rate, with no net change in the concentrations of reactants and products over time. The yield is influenced by reaction conditions such as temperature and pressure, and understanding how to optimize these conditions is essential in industry and research.According to Le Chatelier's principle, changing the reaction conditions can shift the position of equilibrium and thus increase the yield of the desired product. In our methanol synthesis example, the desired equilibrium yield is maximized by choosing a low temperature and a high pressure. This approach is commonly employed in the production of many chemicals to maximize efficiency and profitability.

Standard Enthalpy of Formation

The standard enthalpy of formation (e>e>e>e>e>) is the heat change that accompanies the formation of one mole of a compound from its elements in their standard states. It's a foundational concept in thermochemistry used to calculate the enthalpy change of reactions (e>e>e>e>e>). A negative value of enthalpy indicates that the reaction is exothermic, releasing energy to the surroundings; a positive value means the reaction is endothermic and absorbs energy.For the methanol synthesis reaction, the standard enthalpy of formation of the reactants and product allows us to calculate the overall d>d>d). This calculated value tells us not only the amount of energy involved but also helps dictate, through Le Chatelier's principle, how variations in temperature and pressure can shift the equilibrium to favor the production of methanol. Understanding this concept helps chemists design processes and predict the behavior of reactions under different conditions.