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Chapter 15: Problem 66

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

Expert verified
a) The standard enthalpy change (∆H°) for the reaction is -90.5 kJ/mol. b) To maximize the equilibrium yield of methanol, we would use a low temperature. c) To maximize the equilibrium yield of methanol, we would use a high pressure.

Step by step solution

01

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

To calculate the standard enthalpy change for the given reaction, we'll need to use the standard enthalpies of formation values provided in Appendix C. The equation to find the ∆H° of the reaction is: ∆H° = Σ (n * ∆H°f of products) - Σ (n * ∆H°f of reactants) Here, n represents the stoichiometric coefficients of the products and reactants. From Appendix C, we can find the standard enthalpies of formation: ∆H°f (CO) = -110.5 kJ/mol ∆H°f (H2) = 0 kJ/mol (since it's an element in its standard state) ∆H°f (CH3OH) = -201.0 kJ/mol Using these values, we can calculate the ∆H° for the reaction: ∆H° = (1 * -201.0) - (1 * -110.5) = -90.5 kJ/mol The standard enthalpy change for the reaction is -90.5 kJ/mol.
02

Effect of temperature on equilibrium yield of methanol

Since the reaction is exothermic (∆H° < 0), according to Le Chatelier's Principle, increasing the temperature would favor the endothermic (reverse) reaction and decrease the equilibrium yield of methanol. Therefore, to maximize the equilibrium yield of methanol, we would use a low temperature.
03

Effect of pressure on equilibrium yield of methanol

According to Le Chatelier's Principle, increasing the pressure would favor the side of the reaction with fewer moles of gas. In this case, the reactant side (CO + 2H2) has 3 moles of gas, while the product side (CH3OH) has only 1 mole of gas. So, increasing the pressure would favor the formation of methanol, which has fewer moles of gas. To maximize the equilibrium yield of methanol, we would use a high pressure. In conclusion: a) The standard enthalpy change (∆H°) for the reaction is -90.5 kJ/mol. b) To maximize the equilibrium yield of methanol, we would use a low temperature. c) To maximize the equilibrium yield of methanol, we would use a high pressure.

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Most popular questions from this chapter

Ethene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) reacts with halogens \(\left(\mathrm{X}_{2}\right)\) by the following reaction: $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{X}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{X}_{2}(g) $$ The following figures represent the concentrations at equilibrium at the same temperature when \(\mathrm{X}_{2}\) is \(\mathrm{Cl}_{2}\) (green), \(\mathrm{Br}_{2}\) (brown), and \(\mathrm{I}_{2}\) (purple). List the equilibria from smallest to largest equilibrium constant. [Section 15.3\(]\)

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