Methanol (CH3OH) can be made by the reaction of CO with H2: CO (g) + 2H2 (g) ⇌CH3OH (g) Part A Use thermochemical data in Appendix C in the textbook to calculate ΔH∘ for this reaction. ΔH∘Δ H ∘ = nothing kJ SubmitRequest Answer Part B To maximize the equilibrium yield of methanol, would you use a high or low temperature? To maximize the equilibrium yield of methanol, would you use a high or low temperature? high temprerature low temperature SubmitRequest Answer Part C To maximize the equilibrium yield of methanol, would you use a high or low pressure? To maximize the equilibrium yield of methanol, would you use a high or low pressure? high pressure low pressure SubmitRequest Answer Provide Feedback Next

Question

Chemistry

Isabella Heath

12 April, 17:27

Methanol (CH3OH) can be made by the reaction of CO with H2: CO (g) + 2H2 (g) ⇌CH3OH (g) Part A Use thermochemical data in Appendix C in the textbook to calculate ΔH∘ for this reaction. ΔH∘Δ H ∘ = nothing kJ SubmitRequest Answer Part B To maximize the equilibrium yield of methanol, would you use a high or low temperature? To maximize the equilibrium yield of methanol, would you use a high or low temperature? high temprerature low temperature SubmitRequest Answer Part C To maximize the equilibrium yield of methanol, would you use a high or low pressure? To maximize the equilibrium yield of methanol, would you use a high or low pressure? high pressure low pressure SubmitRequest Answer Provide Feedback Next

+3

Answers (1)

Know the Answer?

Angeline Parks · Answer 1

Part A: ΔH°r = - 90.80 kJ

Part B: Low temperature.

Part C: High pressure

Explanation:

Part A

To calculate the standard enthalpy of reaction (ΔH°r) we can use the standard enthalpies of formation using the following expression:

ΔH°r = ∑n (p).ΔH°f (p) - ∑n (r).ΔH°f (r)

where,

n (i) is the number of moles of reactants and products in the balanced equation.

ΔH°f (i) is the standard enthalpy of formation of reactants in products, which are tabulated.

For the reaction:

CO (g) + 2 H₂ (g) ⇌CH₃OH (g)

The standard enthalpy of reaction is:

ΔH°r = [1 mol x ΔH°f (CH₃OH (g)) ] - [1 mol x ΔH°f (CO (g)) + 2 mol x ΔH°f (H₂ (g)) ]

ΔH°r = [1 mol x (-201.3 kJ/mol) ] - [1 mol x (-110.5 kJ/mol) + 2 mol x 0 kJ/mol]

ΔH°r = - 90.80 kJ

Part B: To maximize the equilibrium yield of methanol, would you use a high or low temperature?

In Part B and C we will use Le Chatelier Principle: if a system at equilibrium undergoes a change, it will shift its equilibrium to counteract such change. This is used in industry to increase the yield of the product.

This reaction is exothermic (ΔH°r is negative). If we lower the temperature, the system will try to increase the temperature by favoring the forward reaction, thus releasing heat.

Part C: To maximize the equilibrium yield of methanol, would you use a high or low pressure?

Considering Le Chatelier Principle, if we increase the pressure, the system will try to decrease it by shifting its equilibrium towards the side where there are fewer moles of gases, in this case, the right side (1 mole of gas vs. 3 moles of gases on the left side).