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C₃H₈O(g) ⇌ C₃H₆O(g) + H₂(g); ΔH = -54 kJ mol⁻¹ At equilibrium at a particular temperature, 10% of the 2-propanol is converted to propanone - VCE - SSCE Chemistry - Question 6 - 2007 - Paper 1
Question 6
C₃H₈O(g) ⇌ C₃H₆O(g) + H₂(g); ΔH = -54 kJ mol⁻¹ At equilibrium at a particular temperature, 10% of the 2-propanol is converted to propanone. In order to increase the... show full transcript
Worked Solution & Example Answer:C₃H₈O(g) ⇌ C₃H₆O(g) + H₂(g); ΔH = -54 kJ mol⁻¹ At equilibrium at a particular temperature, 10% of the 2-propanol is converted to propanone - VCE - SSCE Chemistry - Question 6 - 2007 - Paper 1
Step 1
In order to increase the percentage yield of propanone at equilibrium, you should:
Answer
To determine how to increase the yield of propanone, consider Le Chatelier's principle, which states that a system at equilibrium will adjust to counteract the effects of a change in conditions.
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Exothermic Reaction: The given reaction is exothermic (indicated by the negative ΔH of -54 kJ mol⁻¹), meaning heat can be considered as a product.
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Pressure Changes: Reducing pressure shifts the equilibrium towards the side with more gaseous moles. Analyzing the reaction, there is 1 mole of gas on the right (H₂) and 1 mole of gas on the left (C₃H₈O). Therefore, pressure changes do not affect the equilibrium position significantly in this case.
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Temperature Changes: To maximize the yield of propanone, we should lower the temperature since this is an exothermic reaction. Cooling the system will shift the equilibrium to the right, favoring the production of propanone and hydrogen.
Given this information, the correct choice to increase the yield of propanone is:
A. lower the temperature and raise the pressure.