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25 mL of hydrogen and 18 mL of Iodine when heated in closed container, produced 30.8 mL of HI at equilibrium. Calculate degree of dissociation of HI at same temperature:
- a)0.54
- b)0.245
- c)0.38
- d)0.96
Correct answer is option 'B'. Can you explain this answer?
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25 mL of hydrogen and 18 mL of Iodine when heated in closed container,...
Given:
Volume of hydrogen (V₁) = 25 mL
Volume of iodine (V₂) = 18 mL
Volume of HI produced (V₃) = 30.8 mL
We know that the balanced chemical equation for the reaction is:
H₂ (g) + I₂ (g) ⇌ 2HI (g)
To calculate the degree of dissociation of HI, we need to calculate the initial concentration of H₂ and I₂, as well as the equilibrium concentration of HI.
Calculation:
1. Calculate the number of moles of H₂ and I₂.
- n(H₂) = V₁/22.4 = 25/22.4 = 1.116 mol
- n(I₂) = V₂/22.4 = 18/22.4 = 0.804 mol
2. Assume x moles of HI dissociate at equilibrium. Then the equilibrium concentration of HI is:
- [HI] = (2n(H₂) - x)/(V₁/1000) = (2(1.116) - x)/(25/1000) = 44.64 - 40x
3. Since the initial concentration of H₂ and I₂ is equal, each is 1.116/0.025 = 44.64 mol/L.
4. Using the equilibrium constant expression, we can write:
- Kc = ([HI]²)/([H₂][I₂]) = (44.64 - 40x)²/(44.64)²
5. Substitute the given values in the equation V₃ = n(HI)(RT/P) to find n(HI) and substitute it in the equilibrium constant expression.
- n(HI) = V₃(1 atm)(273 K)/(0.0821 L atm/mol K)(1000 mL/L) = 1.225 mol
- Kc = (2.45 - 2x)²/(1.116)(0.804) = 3.054 - 7.339x + 4x²
6. Solve the quadratic equation to find the value of x:
- 4x² - 7.339x + 1.554 = 0
- x = (7.339 ± √(7.339² - 4(4)(1.554)))/(2(4)) = 0.245 or 0.317
Answer:
The degree of dissociation of HI at the same temperature is 0.245, which is option (b).
Volume of hydrogen (V₁) = 25 mL
Volume of iodine (V₂) = 18 mL
Volume of HI produced (V₃) = 30.8 mL
We know that the balanced chemical equation for the reaction is:
H₂ (g) + I₂ (g) ⇌ 2HI (g)
To calculate the degree of dissociation of HI, we need to calculate the initial concentration of H₂ and I₂, as well as the equilibrium concentration of HI.
Calculation:
1. Calculate the number of moles of H₂ and I₂.
- n(H₂) = V₁/22.4 = 25/22.4 = 1.116 mol
- n(I₂) = V₂/22.4 = 18/22.4 = 0.804 mol
2. Assume x moles of HI dissociate at equilibrium. Then the equilibrium concentration of HI is:
- [HI] = (2n(H₂) - x)/(V₁/1000) = (2(1.116) - x)/(25/1000) = 44.64 - 40x
3. Since the initial concentration of H₂ and I₂ is equal, each is 1.116/0.025 = 44.64 mol/L.
4. Using the equilibrium constant expression, we can write:
- Kc = ([HI]²)/([H₂][I₂]) = (44.64 - 40x)²/(44.64)²
5. Substitute the given values in the equation V₃ = n(HI)(RT/P) to find n(HI) and substitute it in the equilibrium constant expression.
- n(HI) = V₃(1 atm)(273 K)/(0.0821 L atm/mol K)(1000 mL/L) = 1.225 mol
- Kc = (2.45 - 2x)²/(1.116)(0.804) = 3.054 - 7.339x + 4x²
6. Solve the quadratic equation to find the value of x:
- 4x² - 7.339x + 1.554 = 0
- x = (7.339 ± √(7.339² - 4(4)(1.554)))/(2(4)) = 0.245 or 0.317
Answer:
The degree of dissociation of HI at the same temperature is 0.245, which is option (b).
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25 mL of hydrogen and 18 mL of Iodine when heated in closed container, produced 30.8 mL of HI at equilibrium. Calculate degree of dissociation of HI at same temperature:a)0.54b)0.245c)0.38d)0.96Correct answer is option 'B'. Can you explain this answer?
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25 mL of hydrogen and 18 mL of Iodine when heated in closed container, produced 30.8 mL of HI at equilibrium. Calculate degree of dissociation of HI at same temperature:a)0.54b)0.245c)0.38d)0.96Correct answer is option 'B'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about 25 mL of hydrogen and 18 mL of Iodine when heated in closed container, produced 30.8 mL of HI at equilibrium. Calculate degree of dissociation of HI at same temperature:a)0.54b)0.245c)0.38d)0.96Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 25 mL of hydrogen and 18 mL of Iodine when heated in closed container, produced 30.8 mL of HI at equilibrium. Calculate degree of dissociation of HI at same temperature:a)0.54b)0.245c)0.38d)0.96Correct answer is option 'B'. Can you explain this answer?.
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