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Methanol can be produced from the reaction of carbon monoxide and hydrogen, according to the following equation: CO(g) + 2H2(g) ⇌ CH3OH(g) ΔH° = -90 kJ mol⁻¹ Which set of conditions will produce the maximum yield of methanol? A - HSC - SSCE Chemistry - Question 12 - 2019 - Paper 1

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Methanol-can-be-produced-from-the-reaction-of-carbon-monoxide-and-hydrogen,-according-to-the-following-equation:--CO(g)-+-2H2(g)-⇌-CH3OH(g)--ΔH°-=--90-kJ-mol⁻¹--Which-set-of-conditions-will-produce-the-maximum-yield-of-methanol?--A-HSC-SSCE Chemistry-Question 12-2019-Paper 1.png

Methanol can be produced from the reaction of carbon monoxide and hydrogen, according to the following equation: CO(g) + 2H2(g) ⇌ CH3OH(g) ΔH° = -90 kJ mol⁻¹ Whic... show full transcript

Worked Solution & Example Answer:Methanol can be produced from the reaction of carbon monoxide and hydrogen, according to the following equation: CO(g) + 2H2(g) ⇌ CH3OH(g) ΔH° = -90 kJ mol⁻¹ Which set of conditions will produce the maximum yield of methanol? A - HSC - SSCE Chemistry - Question 12 - 2019 - Paper 1

Step 1

Which set of conditions will produce the maximum yield of methanol?

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Answer

To determine the conditions that yield the maximum methanol production, we can apply Le Chatelier's principle and consider the nature of the reaction.

The reaction is exothermic (ΔH°=90 kJ mol1\Delta H° = -90 \text{ kJ mol}^{-1}), indicating that heat is released when methanol is produced. To achieve a higher yield of methanol:

  1. Temperature Consideration: Lowering the temperature favors exothermic reactions. Therefore, lower temperatures would increase methanol production.
  2. Pressure Consideration: The reaction involves a decrease in the number of moles of gas (1 mole of CO and 2 moles of H2 combine to form 1 mole of CH3OH), which means that increasing pressure will favor the formation of methanol.

Combining these two conditions leads us to the conclusion that high pressure and low temperature (option C) will promote the maximum yield of methanol. Hence, the correct answer is C.

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