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At constant temperature in one litre vessel, when the reaction, 2SO3(g)→2SO2(g)+O2(g) is at equilibrium, the SO2 concentration is 0.6 M, initial concentration of SO3 is 1 M. The equilibrium constant is:    
  • a)
    2.7        
  • b)
        1.36            
  • c)
        0.34            
  • d)
        0.675
Correct answer is option 'D'. Can you explain this answer?
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At constant temperature in one litre vessel, when the reaction, 2SO3(g...

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At constant temperature in one litre vessel, when the reaction, 2SO3(g...
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2SO2 + O2 <=> 2SO​3 ____Kc=278

SO3 <=> SO 2 + 1/2O2 _____Kc'=(1/Kc)1/2

Kc'=(1/278)1/2 = 0.0599 = 5.9 x 10-2 =6x10-2


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At constant temperature in one litre vessel, when the reaction, 2SO3(g...
2SO2(g) + O2(g), is allowed to reach equilibrium, it is found that the concentration of SO3(g) is 0.2 M. What are the concentrations of SO2(g) and O2(g) at equilibrium?

To solve this problem, we can use the stoichiometry of the reaction and the equilibrium constant expression.

The balanced equation for the reaction is:
2SO3(g) ↔ 2SO2(g) + O2(g)

The equilibrium constant expression for this reaction is:
Kc = [SO2]^2 * [O2] / [SO3]^2

Given that the concentration of SO3(g) is 0.2 M, we can substitute this value into the equilibrium constant expression:
0.2 = (2[SO2])^2 * [O2] / (0.2)^2

Simplifying the equation, we get:
0.2 = 4[SO2]^2 * [O2] / 0.04

Rearranging the equation, we have:
[SO2]^2 * [O2] = 0.04

Since the volume of the vessel is constant at 1 liter, we can assume that the initial moles of SO3, SO2, and O2 are equal to their concentrations. Let's call this initial concentration x.

Therefore, we have:
x^2 * x = 0.04

Solving for x, we find:
x^3 = 0.04
x = 0.04^(1/3)
x = 0.32 M

So, the concentration of SO2(g) and O2(g) at equilibrium is 0.32 M.
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At constant temperature in one litre vessel, when the reaction, 2SO3(g)→2SO2(g)+O2(g) is at equilibrium, the SO2concentration is 0.6 M, initial concentration of SO3 is 1 M. The equilibrium constant is: a)2.7 b) 1.36 c) 0.34 d) 0.675Correct answer is option 'D'. Can you explain this answer?
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At constant temperature in one litre vessel, when the reaction, 2SO3(g)→2SO2(g)+O2(g) is at equilibrium, the SO2concentration is 0.6 M, initial concentration of SO3 is 1 M. The equilibrium constant is: a)2.7 b) 1.36 c) 0.34 d) 0.675Correct answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about At constant temperature in one litre vessel, when the reaction, 2SO3(g)→2SO2(g)+O2(g) is at equilibrium, the SO2concentration is 0.6 M, initial concentration of SO3 is 1 M. The equilibrium constant is: a)2.7 b) 1.36 c) 0.34 d) 0.675Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At constant temperature in one litre vessel, when the reaction, 2SO3(g)→2SO2(g)+O2(g) is at equilibrium, the SO2concentration is 0.6 M, initial concentration of SO3 is 1 M. The equilibrium constant is: a)2.7 b) 1.36 c) 0.34 d) 0.675Correct answer is option 'D'. Can you explain this answer?.
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