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At certain temperature, 50% of HI is dissociated into H2 and I2, the equilibrium constant is
  • a)
    1.0
  • b)
    3.0
  • c)
    0.5
  • d)
    0.25
Correct answer is option 'D'. Can you explain this answer?
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Understanding the Problem:
At a certain temperature, 50% of HI is dissociated into H2 and I2. The equilibrium constant (Kc) needs to be determined.

Solving the Problem:
- Let's assume the initial concentration of HI is 'x' mol/L.
- At equilibrium, the concentration of H2 and I2 formed will be '0.5x' mol/L each.
- The equilibrium constant (Kc) expression for the reaction is:
Kc = [H2] * [I2] / [HI]
- Substituting the equilibrium concentrations into the expression:
Kc = (0.5x) * (0.5x) / (x - 0.5x)
Kc = 0.25x^2 / 0.5x
Kc = 0.5

Conclusion:
The equilibrium constant (Kc) for the given reaction is 0.5, which corresponds to option 'c'.
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At certain temperature, 50% of HI is dissociated into H2 and I2, the equilibrium constant isa)1.0b)3.0c)0.5d)0.25Correct answer is option 'D'. Can you explain this answer?
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At certain temperature, 50% of HI is dissociated into H2 and I2, the equilibrium constant isa)1.0b)3.0c)0.5d)0.25Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At certain temperature, 50% of HI is dissociated into H2 and I2, the equilibrium constant isa)1.0b)3.0c)0.5d)0.25Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At certain temperature, 50% of HI is dissociated into H2 and I2, the equilibrium constant isa)1.0b)3.0c)0.5d)0.25Correct answer is option 'D'. Can you explain this answer?.
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