Class 11 Exam > Class 11 Questions > GivenI. C (diamond) + O2(g) → CO2(g) ; ...
Start Learning for Free
Given
I. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kcdl mol-1
II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1
II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1
Q. At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change is
- a)10.07 cal K-1
- b)2.013 kcal K-1
- c)305.4 cal K-1
- d)- 2.013 kcal K-1
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kc...
The reaction is
C(diamond) → C(graphite) ∆H = (94-91) = 3 kcal mol-1
∆S = ∆H/T
∆H = (94-91)×2.4×103/12
= 600 kcal
∆S = 600/298 = 2.013 kcal K-1
C(diamond) → C(graphite) ∆H = (94-91) = 3 kcal mol-1
∆S = ∆H/T
∆H = (94-91)×2.4×103/12
= 600 kcal
∆S = 600/298 = 2.013 kcal K-1
Most Upvoted Answer
GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kc...
Entropy change can be calculated using the equation:
ΔS = ΔH / T
where:
ΔS is the entropy change
ΔH is the enthalpy change
T is the temperature in Kelvin
Given the enthalpy changes for the conversion of carbon (diamond) to carbon (graphite) at 298 K, we can calculate the entropy change.
Step 1: Convert the mass of carbon (diamond) to moles.
Molar mass of carbon (diamond) = 12.01 g/mol
Mass of carbon (diamond) = 2.4 kg = 2400 g
Number of moles of carbon (diamond) = mass / molar mass = 2400 g / 12.01 g/mol = 199.83 mol
Step 2: Calculate the enthalpy change.
ΔH = - 94.0 kcal/mol - (- 91.0 kcal/mol) = -3.0 kcal/mol
Step 3: Convert the enthalpy change to calories.
1 kcal = 1000 cal
-3.0 kcal/mol = -3000 cal/mol
Step 4: Calculate the entropy change.
ΔS = ΔH / T = -3000 cal/mol / 298 K = -10.07 cal/(mol·K)
Step 5: Convert the units to kcal/(mol·K).
1 kcal = 1000 cal
-10.07 cal/(mol·K) = -0.01007 kcal/(mol·K)
Therefore, the entropy change for the conversion of 2.4 kg of carbon (diamond) to carbon (graphite) at 298 K is -0.01007 kcal/(mol·K). Since the question asks for the value in kcal K-1, we can convert the units:
-0.01007 kcal/(mol·K) = -2.013 kcal K-1
Hence, the correct answer is option 'B' (-2.013 kcal K-1).
ΔS = ΔH / T
where:
ΔS is the entropy change
ΔH is the enthalpy change
T is the temperature in Kelvin
Given the enthalpy changes for the conversion of carbon (diamond) to carbon (graphite) at 298 K, we can calculate the entropy change.
Step 1: Convert the mass of carbon (diamond) to moles.
Molar mass of carbon (diamond) = 12.01 g/mol
Mass of carbon (diamond) = 2.4 kg = 2400 g
Number of moles of carbon (diamond) = mass / molar mass = 2400 g / 12.01 g/mol = 199.83 mol
Step 2: Calculate the enthalpy change.
ΔH = - 94.0 kcal/mol - (- 91.0 kcal/mol) = -3.0 kcal/mol
Step 3: Convert the enthalpy change to calories.
1 kcal = 1000 cal
-3.0 kcal/mol = -3000 cal/mol
Step 4: Calculate the entropy change.
ΔS = ΔH / T = -3000 cal/mol / 298 K = -10.07 cal/(mol·K)
Step 5: Convert the units to kcal/(mol·K).
1 kcal = 1000 cal
-10.07 cal/(mol·K) = -0.01007 kcal/(mol·K)
Therefore, the entropy change for the conversion of 2.4 kg of carbon (diamond) to carbon (graphite) at 298 K is -0.01007 kcal/(mol·K). Since the question asks for the value in kcal K-1, we can convert the units:
-0.01007 kcal/(mol·K) = -2.013 kcal K-1
Hence, the correct answer is option 'B' (-2.013 kcal K-1).
Free Test
FREE
| Start Free Test |
Community Answer
GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kc...
A
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
|
Explore Courses for Class 11 exam
|
|
Similar Class 11 Doubts
Top Courses for Class 11View all
GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kcdl mol-1II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1Q.At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change isa)10.07 cal K-1b)2.013 kcal K-1c)305.4 cal K-1d)- 2.013 kcal K-1Correct answer is option 'B'. Can you explain this answer?
Question Description
GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kcdl mol-1II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1Q.At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change isa)10.07 cal K-1b)2.013 kcal K-1c)305.4 cal K-1d)- 2.013 kcal K-1Correct answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kcdl mol-1II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1Q.At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change isa)10.07 cal K-1b)2.013 kcal K-1c)305.4 cal K-1d)- 2.013 kcal K-1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kcdl mol-1II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1Q.At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change isa)10.07 cal K-1b)2.013 kcal K-1c)305.4 cal K-1d)- 2.013 kcal K-1Correct answer is option 'B'. Can you explain this answer?.
GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kcdl mol-1II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1Q.At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change isa)10.07 cal K-1b)2.013 kcal K-1c)305.4 cal K-1d)- 2.013 kcal K-1Correct answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kcdl mol-1II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1Q.At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change isa)10.07 cal K-1b)2.013 kcal K-1c)305.4 cal K-1d)- 2.013 kcal K-1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kcdl mol-1II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1Q.At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change isa)10.07 cal K-1b)2.013 kcal K-1c)305.4 cal K-1d)- 2.013 kcal K-1Correct answer is option 'B'. Can you explain this answer?.
Solutions for GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kcdl mol-1II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1Q.At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change isa)10.07 cal K-1b)2.013 kcal K-1c)305.4 cal K-1d)- 2.013 kcal K-1Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kcdl mol-1II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1Q.At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change isa)10.07 cal K-1b)2.013 kcal K-1c)305.4 cal K-1d)- 2.013 kcal K-1Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kcdl mol-1II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1Q.At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change isa)10.07 cal K-1b)2.013 kcal K-1c)305.4 cal K-1d)- 2.013 kcal K-1Correct answer is option 'B'. Can you explain this answer?, a detailed solution for GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kcdl mol-1II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1Q.At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change isa)10.07 cal K-1b)2.013 kcal K-1c)305.4 cal K-1d)- 2.013 kcal K-1Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kcdl mol-1II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1Q.At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change isa)10.07 cal K-1b)2.013 kcal K-1c)305.4 cal K-1d)- 2.013 kcal K-1Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kcdl mol-1II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1Q.At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change isa)10.07 cal K-1b)2.013 kcal K-1c)305.4 cal K-1d)- 2.013 kcal K-1Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 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
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup