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1 mole of H2 and 2 moles of I2 are taken initially in a 2 litre vessel. If the number of moles of H2 at equilibrium is 0.2, the number of moles of I2 and HI, respectively at equilibrium would be
- a)1.2, 1.6
- b)1.8, 1.0
- c)0.4, 2.4
- d)0.8, 2.0
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
1 mole of H2 and 2 moles of I2 are taken initially in a 2 litre vessel...
H2 + I2 ---> 2HI
at initial
1/2 mole/l of H2
2/2 mole/l of I2
and zero mole/l
now at equilibrium
(0.5-x)mole/l of H2
(1-x) mole/l of I2
and (1-2x)mole/l
of HI
here we see limiting reagents is H2 hence
concentration of HI is related to H2 not I2
now a/c
0.5-x=0.2/2=0.1
x=0.4mole/l
hence I2 concentration is 1-0.4=0.6mole/l
hence no of mole of I2
=0.6x2=1.2mole
now concentration of HI=1-2(0.1)=0.8
because concentration of H2=0.1mole/l
now no of mole of HI=0.8x2=1.6
Most Upvoted Answer
1 mole of H2 and 2 moles of I2 are taken initially in a 2 litre vessel...
Given Information:
- Initial moles of H2 = 1 mole
- Initial moles of I2 = 2 moles
- Total volume of the vessel = 2 liters
Explanation:
The reaction between H2 and I2 can be represented as follows:
H2(g) + I2(g) ⇌ 2HI(g)
The balanced equation shows that 1 mole of H2 reacts with 1 mole of I2 to form 2 moles of HI. Therefore, the stoichiometry of the reaction is 1:1:2.
Step 1: Calculate the initial concentrations of H2 and I2:
- Initial concentration of H2 = moles of H2 / volume of vessel = 1 mole / 2 L = 0.5 M
- Initial concentration of I2 = moles of I2 / volume of vessel = 2 moles / 2 L = 1 M
Step 2: Use the stoichiometry of the reaction to determine the change in concentrations:
Since the stoichiometry of the reaction is 1:1:2, if x moles of H2 react, x moles of I2 will also react, and 2x moles of HI will be formed.
Step 3: Express the equilibrium concentrations in terms of the initial concentrations and the change in concentrations:
- Equilibrium concentration of H2 = Initial concentration of H2 - Change in concentration of H2 = 0.5 M - x
- Equilibrium concentration of I2 = Initial concentration of I2 - Change in concentration of I2 = 1 M - x
- Equilibrium concentration of HI = Change in concentration of HI = 2x
Step 4: Use the given information to set up an equation:
The given information states that the equilibrium concentration of H2 is 0.2 moles. Therefore, we can set up the following equation:
0.2 = 0.5 - x
Step 5: Solve the equation to find the value of x:
0.2 = 0.5 - x
x = 0.5 - 0.2
x = 0.3
Step 6: Calculate the equilibrium concentrations:
- Equilibrium concentration of H2 = 0.5 - x = 0.5 - 0.3 = 0.2 M
- Equilibrium concentration of I2 = 1 - x = 1 - 0.3 = 0.7 M
- Equilibrium concentration of HI = 2x = 2 * 0.3 = 0.6 M
Therefore, at equilibrium, the number of moles of I2 and HI are 0.7 moles and 0.6 moles, respectively.
Conclusion:
The correct answer is option A) 1.2 moles of I2 and 1.6 moles of HI at equilibrium.
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1 mole of H2 and 2 moles of I2 are taken initially in a 2 litre vessel. If the number of moles of H2 at equilibrium is 0.2, the number of moles of I2 and HI, respectively at equilibrium would bea)1.2, 1.6b)1.8, 1.0c)0.4, 2.4d)0.8, 2.0Correct answer is option 'A'. Can you explain this answer?
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1 mole of H2 and 2 moles of I2 are taken initially in a 2 litre vessel. If the number of moles of H2 at equilibrium is 0.2, the number of moles of I2 and HI, respectively at equilibrium would bea)1.2, 1.6b)1.8, 1.0c)0.4, 2.4d)0.8, 2.0Correct answer is option 'A'. 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 1 mole of H2 and 2 moles of I2 are taken initially in a 2 litre vessel. If the number of moles of H2 at equilibrium is 0.2, the number of moles of I2 and HI, respectively at equilibrium would bea)1.2, 1.6b)1.8, 1.0c)0.4, 2.4d)0.8, 2.0Correct answer is option 'A'. 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 1 mole of H2 and 2 moles of I2 are taken initially in a 2 litre vessel. If the number of moles of H2 at equilibrium is 0.2, the number of moles of I2 and HI, respectively at equilibrium would bea)1.2, 1.6b)1.8, 1.0c)0.4, 2.4d)0.8, 2.0Correct answer is option 'A'. Can you explain this answer?.
1 mole of H2 and 2 moles of I2 are taken initially in a 2 litre vessel. If the number of moles of H2 at equilibrium is 0.2, the number of moles of I2 and HI, respectively at equilibrium would bea)1.2, 1.6b)1.8, 1.0c)0.4, 2.4d)0.8, 2.0Correct answer is option 'A'. 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 1 mole of H2 and 2 moles of I2 are taken initially in a 2 litre vessel. If the number of moles of H2 at equilibrium is 0.2, the number of moles of I2 and HI, respectively at equilibrium would bea)1.2, 1.6b)1.8, 1.0c)0.4, 2.4d)0.8, 2.0Correct answer is option 'A'. 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 1 mole of H2 and 2 moles of I2 are taken initially in a 2 litre vessel. If the number of moles of H2 at equilibrium is 0.2, the number of moles of I2 and HI, respectively at equilibrium would bea)1.2, 1.6b)1.8, 1.0c)0.4, 2.4d)0.8, 2.0Correct answer is option 'A'. Can you explain this answer?.
Solutions for 1 mole of H2 and 2 moles of I2 are taken initially in a 2 litre vessel. If the number of moles of H2 at equilibrium is 0.2, the number of moles of I2 and HI, respectively at equilibrium would bea)1.2, 1.6b)1.8, 1.0c)0.4, 2.4d)0.8, 2.0Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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ample number of questions to practice 1 mole of H2 and 2 moles of I2 are taken initially in a 2 litre vessel. If the number of moles of H2 at equilibrium is 0.2, the number of moles of I2 and HI, respectively at equilibrium would bea)1.2, 1.6b)1.8, 1.0c)0.4, 2.4d)0.8, 2.0Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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