NEET Exam > NEET Questions > 0.96g HI were heated to attain equilibrium. 2...
Start Learning for Free
0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilibrium mixture, on reaction requires 15ml of M/10 Na2S2O3 solution. Calculate the degree of dissociation of HI + I2 Na2S2O3 = Na2S4O6 + NaI(unbalanced)?
Verified Answer
0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilib...
This question is part of UPSC exam. View all NEET courses
Most Upvoted Answer
0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilib...
Given:
- Initial moles of HI = 0.96g
- Reaction: 2HI ⇌ H2 + I2
- Equilibrium mixture requires 15ml of M/10 Na2S2O3 solution
To find:
- Degree of dissociation of HI
Approach:
1. Calculate the moles of HI initially.
2. Use the stoichiometry of the reaction to determine the moles of I2 formed.
3. Calculate the moles of Na2S2O3 required to react with I2.
4. Use the volume and concentration of Na2S2O3 to calculate the moles of Na2S2O3 used.
5. Determine the moles of I2 reacted based on the moles of Na2S2O3 used.
6. Use the stoichiometry of the reaction to determine the moles of HI reacted.
7. Calculate the degree of dissociation of HI.
Calculation:
1. Moles of HI initially = (0.96g)/(molar mass of HI)
2. Moles of I2 formed = (moles of HI initially)/2 (according to the stoichiometry of the reaction)
3. Moles of Na2S2O3 required to react with I2 = (moles of I2 formed)/(stoichiometric coefficient of I2 in the reaction)
4. Moles of Na2S2O3 used = (volume of Na2S2O3 solution in liters) × (concentration of Na2S2O3 in moles/liter)
5. Moles of I2 reacted = (moles of Na2S2O3 used) × (stoichiometric coefficient of I2 in the reaction)
6. Moles of HI reacted = 2 × (moles of I2 reacted) (according to the stoichiometry of the reaction)
7. Degree of dissociation of HI = (moles of HI reacted)/(moles of HI initially)
Explanation:
The given reaction is the dissociation of 2HI into H2 and I2. To calculate the degree of dissociation of HI, we need to determine the amount of HI that dissociated and reacted to form I2.
First, we calculate the moles of HI initially by dividing the given mass of HI by its molar mass. Then, using the stoichiometry of the reaction, we determine the moles of I2 formed which is half the moles of HI initially.
Next, we calculate the moles of Na2S2O3 required to react with the I2 formed. This is done by dividing the moles of I2 by the stoichiometric coefficient of I2 in the reaction.
To find the moles of Na2S2O3 used, we multiply the volume of the Na2S2O3 solution (given as 15ml) by its concentration (given as M/10). This gives us the moles of Na2S2O3 used in the reaction.
Using the stoichiometry of the reaction, we determine the moles of I2 reacted based on the moles of Na2S2O3 used.
Finally, we calculate the moles of HI reacted by multiplying the moles of I2 reacted by 2 (according to the sto
- Initial moles of HI = 0.96g
- Reaction: 2HI ⇌ H2 + I2
- Equilibrium mixture requires 15ml of M/10 Na2S2O3 solution
To find:
- Degree of dissociation of HI
Approach:
1. Calculate the moles of HI initially.
2. Use the stoichiometry of the reaction to determine the moles of I2 formed.
3. Calculate the moles of Na2S2O3 required to react with I2.
4. Use the volume and concentration of Na2S2O3 to calculate the moles of Na2S2O3 used.
5. Determine the moles of I2 reacted based on the moles of Na2S2O3 used.
6. Use the stoichiometry of the reaction to determine the moles of HI reacted.
7. Calculate the degree of dissociation of HI.
Calculation:
1. Moles of HI initially = (0.96g)/(molar mass of HI)
2. Moles of I2 formed = (moles of HI initially)/2 (according to the stoichiometry of the reaction)
3. Moles of Na2S2O3 required to react with I2 = (moles of I2 formed)/(stoichiometric coefficient of I2 in the reaction)
4. Moles of Na2S2O3 used = (volume of Na2S2O3 solution in liters) × (concentration of Na2S2O3 in moles/liter)
5. Moles of I2 reacted = (moles of Na2S2O3 used) × (stoichiometric coefficient of I2 in the reaction)
6. Moles of HI reacted = 2 × (moles of I2 reacted) (according to the stoichiometry of the reaction)
7. Degree of dissociation of HI = (moles of HI reacted)/(moles of HI initially)
Explanation:
The given reaction is the dissociation of 2HI into H2 and I2. To calculate the degree of dissociation of HI, we need to determine the amount of HI that dissociated and reacted to form I2.
First, we calculate the moles of HI initially by dividing the given mass of HI by its molar mass. Then, using the stoichiometry of the reaction, we determine the moles of I2 formed which is half the moles of HI initially.
Next, we calculate the moles of Na2S2O3 required to react with the I2 formed. This is done by dividing the moles of I2 by the stoichiometric coefficient of I2 in the reaction.
To find the moles of Na2S2O3 used, we multiply the volume of the Na2S2O3 solution (given as 15ml) by its concentration (given as M/10). This gives us the moles of Na2S2O3 used in the reaction.
Using the stoichiometry of the reaction, we determine the moles of I2 reacted based on the moles of Na2S2O3 used.
Finally, we calculate the moles of HI reacted by multiplying the moles of I2 reacted by 2 (according to the sto
Attention NEET Students!
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.
|
|
Explore Courses for NEET exam
|
|
Similar NEET Doubts
Top Courses for NEETView all
0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilibrium mixture, on reaction requires 15ml of M/10 Na2S2O3 solution. Calculate the degree of dissociation of HI + I2 Na2S2O3 = Na2S4O6 + NaI(unbalanced)?
Question Description
0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilibrium mixture, on reaction requires 15ml of M/10 Na2S2O3 solution. Calculate the degree of dissociation of HI + I2 Na2S2O3 = Na2S4O6 + NaI(unbalanced)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilibrium mixture, on reaction requires 15ml of M/10 Na2S2O3 solution. Calculate the degree of dissociation of HI + I2 Na2S2O3 = Na2S4O6 + NaI(unbalanced)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilibrium mixture, on reaction requires 15ml of M/10 Na2S2O3 solution. Calculate the degree of dissociation of HI + I2 Na2S2O3 = Na2S4O6 + NaI(unbalanced)?.
0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilibrium mixture, on reaction requires 15ml of M/10 Na2S2O3 solution. Calculate the degree of dissociation of HI + I2 Na2S2O3 = Na2S4O6 + NaI(unbalanced)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilibrium mixture, on reaction requires 15ml of M/10 Na2S2O3 solution. Calculate the degree of dissociation of HI + I2 Na2S2O3 = Na2S4O6 + NaI(unbalanced)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilibrium mixture, on reaction requires 15ml of M/10 Na2S2O3 solution. Calculate the degree of dissociation of HI + I2 Na2S2O3 = Na2S4O6 + NaI(unbalanced)?.
Solutions for 0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilibrium mixture, on reaction requires 15ml of M/10 Na2S2O3 solution. Calculate the degree of dissociation of HI + I2 Na2S2O3 = Na2S4O6 + NaI(unbalanced)? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of 0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilibrium mixture, on reaction requires 15ml of M/10 Na2S2O3 solution. Calculate the degree of dissociation of HI + I2 Na2S2O3 = Na2S4O6 + NaI(unbalanced)? defined & explained in the simplest way possible. Besides giving the explanation of
0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilibrium mixture, on reaction requires 15ml of M/10 Na2S2O3 solution. Calculate the degree of dissociation of HI + I2 Na2S2O3 = Na2S4O6 + NaI(unbalanced)?, a detailed solution for 0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilibrium mixture, on reaction requires 15ml of M/10 Na2S2O3 solution. Calculate the degree of dissociation of HI + I2 Na2S2O3 = Na2S4O6 + NaI(unbalanced)? has been provided alongside types of 0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilibrium mixture, on reaction requires 15ml of M/10 Na2S2O3 solution. Calculate the degree of dissociation of HI + I2 Na2S2O3 = Na2S4O6 + NaI(unbalanced)? theory, EduRev gives you an
ample number of questions to practice 0.96g HI were heated to attain equilibrium. 2HI = H2 + I2. The equilibrium mixture, on reaction requires 15ml of M/10 Na2S2O3 solution. Calculate the degree of dissociation of HI + I2 Na2S2O3 = Na2S4O6 + NaI(unbalanced)? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup