Class 11 Exam > Class 11 Questions > GivenI. C (diamond) + O2(g) → CO2(g) ; ...
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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?
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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).
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GivenI. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kc...
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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?
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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.
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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.
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