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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
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

01

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.
02

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.
03

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.

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Most popular questions from this chapter

Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: $2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)$. (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibriumconstant expression in terms of molarities for the reaction, using (solv) to indicate solvation.

At \(2000^{\circ} \mathrm{C}\), the equilibrium constant for the reaction $2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$ is \(K_{c}=2.4 \times 10^{3} .\) If the initial concentration of NO is \(0.250 \mathrm{M}\), what are the equilibrium concentrations of \(\mathrm{NO}, \mathrm{N}_{2}\), and \(\mathrm{O}_{2}\) ?

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually $\left.\mathrm{H}_{2} \mathrm{SO}_{4}\right)$. A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{solv})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{solv}) & \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\text { solv })+\mathrm{H}_{2} \mathrm{O}(\text { solv }) \end{aligned}$$ where "(solv)" indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is 6.68 . A pharmaceutical chemist makes up $15.0 \mathrm{~L}\( of a solution that is initially \)0.275 \mathrm{M}$ in acetic acid and \(3.85 \mathrm{M}\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

Calculate \(K_{c}\) at \(900 \mathrm{~K}\) for $2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\( if \)K_{p}=0.0572$ at this temperature.

A 5.37 -g sample of \(\mathrm{SO}_{3}\) is placed in a 5.00-L container and heated to \(1000 \mathrm{~K}\). The \(\mathrm{SO}_{3}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) : $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ At equilibrium, the total pressure in the container is 157 \(\mathrm{kPa}\). Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at \(1100 \mathrm{~K}\).
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