Chemistry Exam > Chemistry Questions > For the reactionH2(g)+I2(g)⇌2HI (g) at ...
Start Learning for Free
For the reaction H2(g)+I2(g)⇌2HI (g) at 721 K
value of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value of KpKp under the same conditions will be
value of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value of KpKp under the same conditions will be
- a)0.05
- b)50
- c)0.1
- d)25 RT
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium...
The equilibrium reaction is H2(g)+I2(g)⇔2HI(g).
The relationship between Kp and Kc is Kp=Kc(RT)Δn.
For the equilibrium reaction, Δn=2−(1+1)=0.
Hence, Kp=Kc=50.
The relationship between Kp and Kc is Kp=Kc(RT)Δn.
For the equilibrium reaction, Δn=2−(1+1)=0.
Hence, Kp=Kc=50.
Most Upvoted Answer
For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium...
Given information:
- Reaction: H2(g) + I2(g) ⇌ 2HI(g)
- Temperature: 721 K
- Equilibrium constant (Kc) at 721 K: 50 (given)
- Equilibrium concentrations: [H2] = [I2] = 0.5 M
To find: Value of equilibrium constant (Kp) under the same conditions
Step-by-step solution:
1. Definition of Kp: Kp is the equilibrium constant expressed in terms of partial pressures of the gases in the reaction mixture.
2. Relationship between Kc and Kp: For any gaseous reaction, the equilibrium constant Kp is related to the equilibrium constant Kc by the following equation:
Kp = Kc(RT)Δn
where R is the gas constant (0.0821 L atm mol^-1 K^-1), T is the temperature in Kelvin, and Δn is the difference between the total number of moles of gaseous products and the total number of moles of gaseous reactants. In this case, Δn = (2-2) = 0.
3. Calculation of Kp: Substituting the given values in the above equation, we get:
Kp = Kc(RT)Δn
Kp = 50(0.0821 x 721)^0
Kp = 50(59.32)
Kp = 2966
4. Answer: Therefore, the value of equilibrium constant (Kp) under the same conditions is 2966, which corresponds to option B.
- Reaction: H2(g) + I2(g) ⇌ 2HI(g)
- Temperature: 721 K
- Equilibrium constant (Kc) at 721 K: 50 (given)
- Equilibrium concentrations: [H2] = [I2] = 0.5 M
To find: Value of equilibrium constant (Kp) under the same conditions
Step-by-step solution:
1. Definition of Kp: Kp is the equilibrium constant expressed in terms of partial pressures of the gases in the reaction mixture.
2. Relationship between Kc and Kp: For any gaseous reaction, the equilibrium constant Kp is related to the equilibrium constant Kc by the following equation:
Kp = Kc(RT)Δn
where R is the gas constant (0.0821 L atm mol^-1 K^-1), T is the temperature in Kelvin, and Δn is the difference between the total number of moles of gaseous products and the total number of moles of gaseous reactants. In this case, Δn = (2-2) = 0.
3. Calculation of Kp: Substituting the given values in the above equation, we get:
Kp = Kc(RT)Δn
Kp = 50(0.0821 x 721)^0
Kp = 50(59.32)
Kp = 2966
4. Answer: Therefore, the value of equilibrium constant (Kp) under the same conditions is 2966, which corresponds to option B.
Free Test
FREE
| Start Free Test |
Community Answer
For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium...
|
Explore Courses for Chemistry exam
|
|
Similar Chemistry Doubts
For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value ofKpKpunder the same conditions will bea)0.05b)50c)0.1d)25 RTCorrect answer is option 'B'. Can you explain this answer?
Question Description
For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value ofKpKpunder the same conditions will bea)0.05b)50c)0.1d)25 RTCorrect 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 For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value ofKpKpunder the same conditions will bea)0.05b)50c)0.1d)25 RTCorrect 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 For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value ofKpKpunder the same conditions will bea)0.05b)50c)0.1d)25 RTCorrect answer is option 'B'. Can you explain this answer?.
For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value ofKpKpunder the same conditions will bea)0.05b)50c)0.1d)25 RTCorrect 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 For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value ofKpKpunder the same conditions will bea)0.05b)50c)0.1d)25 RTCorrect 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 For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value ofKpKpunder the same conditions will bea)0.05b)50c)0.1d)25 RTCorrect answer is option 'B'. Can you explain this answer?.
Solutions for For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value ofKpKpunder the same conditions will bea)0.05b)50c)0.1d)25 RTCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
Download more important topics, notes, lectures and mock test series for Chemistry Exam by signing up for free.
Here you can find the meaning of For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value ofKpKpunder the same conditions will bea)0.05b)50c)0.1d)25 RTCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value ofKpKpunder the same conditions will bea)0.05b)50c)0.1d)25 RTCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value ofKpKpunder the same conditions will bea)0.05b)50c)0.1d)25 RTCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value ofKpKpunder the same conditions will bea)0.05b)50c)0.1d)25 RTCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice For the reactionH2(g)+I2(g)⇌2HI (g) at 721 Kvalue of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value ofKpKpunder the same conditions will bea)0.05b)50c)0.1d)25 RTCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Chemistry tests.
|
|
Explore Courses for Chemistry 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