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A mixture initially containing 2 mol of CO and 2 mol of H2 comes to equilibrium with methanol, CH3OH, as the product of the reaction CO (g) + 2H2(g) → CH3OH (g). At equilibrium the mixture will contain
- a)2 mol of methanol
- b)more than 1 mol but less than 2 mol of methanol
- c)1 mol of methanol
- d)less than 1 mol of methanol
Correct answer is option 'D'. Can you explain this answer?
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A mixture initially containing 2 mol of CO and 2 mol of H2 comes to eq...
Equilibrium Calculation for CO + 2H2 ↔ CH3OH
Initial mole of CO = 2
Initial mole of H2 = 2
Initial mole of CH3OH = 0
Let the reaction quotient, Qc, be x. Then the equilibrium constant, Kc, can be calculated as follows:
Kc = [CH3OH]^1 / [CO]^1 [H2]^2
At equilibrium, let the mole of CH3OH be x.
So, [CH3OH] = x
[CO] = 2 - x
[H2] = 2 - 2x
Substituting these values in the equilibrium constant expression, we get:
Kc = x / (2 - x)(2 - 2x)^2
Since the equilibrium constant is a constant value at a given temperature, we can use it to calculate the value of x at equilibrium.
Kc = 0.209 (at 298 K)
Therefore, x = Kc (2 - x)(2 - 2x)^2
Solving this equation, we get x = 0.5
This means that at equilibrium, there will be less than 1 mole of CH3OH since the initial mole of CO is greater than the mole of CH3OH at equilibrium. Therefore, the correct answer is option D.
Initial mole of CO = 2
Initial mole of H2 = 2
Initial mole of CH3OH = 0
Let the reaction quotient, Qc, be x. Then the equilibrium constant, Kc, can be calculated as follows:
Kc = [CH3OH]^1 / [CO]^1 [H2]^2
At equilibrium, let the mole of CH3OH be x.
So, [CH3OH] = x
[CO] = 2 - x
[H2] = 2 - 2x
Substituting these values in the equilibrium constant expression, we get:
Kc = x / (2 - x)(2 - 2x)^2
Since the equilibrium constant is a constant value at a given temperature, we can use it to calculate the value of x at equilibrium.
Kc = 0.209 (at 298 K)
Therefore, x = Kc (2 - x)(2 - 2x)^2
Solving this equation, we get x = 0.5
This means that at equilibrium, there will be less than 1 mole of CH3OH since the initial mole of CO is greater than the mole of CH3OH at equilibrium. Therefore, the correct answer is option D.
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A mixture initially containing 2 mol of CO and 2 mol of H2 comes to equilibrium with methanol, CH3OH, as the product of the reaction CO (g) + 2H2(g) CH3OH (g). At equilibrium the mixture will containa)2 mol of methanolb)more than 1 mol but less than 2 mol of methanolc)1 mol of methanold)less than 1 mol of methanolCorrect answer is option 'D'. Can you explain this answer?
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A mixture initially containing 2 mol of CO and 2 mol of H2 comes to equilibrium with methanol, CH3OH, as the product of the reaction CO (g) + 2H2(g) CH3OH (g). At equilibrium the mixture will containa)2 mol of methanolb)more than 1 mol but less than 2 mol of methanolc)1 mol of methanold)less than 1 mol of methanolCorrect answer is option 'D'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A mixture initially containing 2 mol of CO and 2 mol of H2 comes to equilibrium with methanol, CH3OH, as the product of the reaction CO (g) + 2H2(g) CH3OH (g). At equilibrium the mixture will containa)2 mol of methanolb)more than 1 mol but less than 2 mol of methanolc)1 mol of methanold)less than 1 mol of methanolCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mixture initially containing 2 mol of CO and 2 mol of H2 comes to equilibrium with methanol, CH3OH, as the product of the reaction CO (g) + 2H2(g) CH3OH (g). At equilibrium the mixture will containa)2 mol of methanolb)more than 1 mol but less than 2 mol of methanolc)1 mol of methanold)less than 1 mol of methanolCorrect answer is option 'D'. Can you explain this answer?.
A mixture initially containing 2 mol of CO and 2 mol of H2 comes to equilibrium with methanol, CH3OH, as the product of the reaction CO (g) + 2H2(g) CH3OH (g). At equilibrium the mixture will containa)2 mol of methanolb)more than 1 mol but less than 2 mol of methanolc)1 mol of methanold)less than 1 mol of methanolCorrect answer is option 'D'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A mixture initially containing 2 mol of CO and 2 mol of H2 comes to equilibrium with methanol, CH3OH, as the product of the reaction CO (g) + 2H2(g) CH3OH (g). At equilibrium the mixture will containa)2 mol of methanolb)more than 1 mol but less than 2 mol of methanolc)1 mol of methanold)less than 1 mol of methanolCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mixture initially containing 2 mol of CO and 2 mol of H2 comes to equilibrium with methanol, CH3OH, as the product of the reaction CO (g) + 2H2(g) CH3OH (g). At equilibrium the mixture will containa)2 mol of methanolb)more than 1 mol but less than 2 mol of methanolc)1 mol of methanold)less than 1 mol of methanolCorrect answer is option 'D'. Can you explain this answer?.
Solutions for A mixture initially containing 2 mol of CO and 2 mol of H2 comes to equilibrium with methanol, CH3OH, as the product of the reaction CO (g) + 2H2(g) CH3OH (g). At equilibrium the mixture will containa)2 mol of methanolb)more than 1 mol but less than 2 mol of methanolc)1 mol of methanold)less than 1 mol of methanolCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM.
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A mixture initially containing 2 mol of CO and 2 mol of H2 comes to equilibrium with methanol, CH3OH, as the product of the reaction CO (g) + 2H2(g) CH3OH (g). At equilibrium the mixture will containa)2 mol of methanolb)more than 1 mol but less than 2 mol of methanolc)1 mol of methanold)less than 1 mol of methanolCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A mixture initially containing 2 mol of CO and 2 mol of H2 comes to equilibrium with methanol, CH3OH, as the product of the reaction CO (g) + 2H2(g) CH3OH (g). At equilibrium the mixture will containa)2 mol of methanolb)more than 1 mol but less than 2 mol of methanolc)1 mol of methanold)less than 1 mol of methanolCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A mixture initially containing 2 mol of CO and 2 mol of H2 comes to equilibrium with methanol, CH3OH, as the product of the reaction CO (g) + 2H2(g) CH3OH (g). At equilibrium the mixture will containa)2 mol of methanolb)more than 1 mol but less than 2 mol of methanolc)1 mol of methanold)less than 1 mol of methanolCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A mixture initially containing 2 mol of CO and 2 mol of H2 comes to equilibrium with methanol, CH3OH, as the product of the reaction CO (g) + 2H2(g) CH3OH (g). At equilibrium the mixture will containa)2 mol of methanolb)more than 1 mol but less than 2 mol of methanolc)1 mol of methanold)less than 1 mol of methanolCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
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