NEET Exam > NEET Questions > The enthalpy of combustion of C(graphite) and...
Start Learning for Free
The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 and -395.4 KJ mol⁻1 respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol⁻1 is
- a)-1.9
- b)-788.9
- c)1.9
- d)788.9
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 a...
Free Test
FREE
| Start Free Test |
Community Answer
The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 a...
Enthalpy of Combustion of C(graphite) and C(diamond):
The enthalpy of combustion is the heat energy released when a substance undergoes complete combustion. It is usually measured in kilojoules per mole (kJ/mol). In this question, we are given the enthalpy of combustion for C(graphite) and C(diamond) as -393.5 and -395.4 kJ/mol, respectively.
Enthalpy of Conversion of C(graphite) to C(diamond):
The enthalpy of conversion refers to the heat energy absorbed or released when a substance undergoes a change in its physical or chemical state. In this case, we are interested in the enthalpy change when C(graphite) is converted to C(diamond).
To calculate the enthalpy of conversion, we can subtract the enthalpy of combustion of C(diamond) from the enthalpy of combustion of C(graphite).
ΔH = H(C(diamond)) - H(C(graphite))
ΔH = -395.4 kJ/mol - (-393.5 kJ/mol)
ΔH = -395.4 kJ/mol + 393.5 kJ/mol
ΔH = -1.9 kJ/mol
Therefore, the enthalpy of conversion of C(graphite) to C(diamond) is -1.9 kJ/mol.
Explanation:
The negative sign indicates that the conversion of C(graphite) to C(diamond) is an exothermic reaction, meaning it releases heat energy. In this case, the enthalpy change is -1.9 kJ/mol, indicating that the reaction releases 1.9 kJ of heat energy per mole of carbon converted.
Option C, which states the enthalpy of conversion as 1.9 kJ/mol, is the correct answer.
The enthalpy of combustion is the heat energy released when a substance undergoes complete combustion. It is usually measured in kilojoules per mole (kJ/mol). In this question, we are given the enthalpy of combustion for C(graphite) and C(diamond) as -393.5 and -395.4 kJ/mol, respectively.
Enthalpy of Conversion of C(graphite) to C(diamond):
The enthalpy of conversion refers to the heat energy absorbed or released when a substance undergoes a change in its physical or chemical state. In this case, we are interested in the enthalpy change when C(graphite) is converted to C(diamond).
To calculate the enthalpy of conversion, we can subtract the enthalpy of combustion of C(diamond) from the enthalpy of combustion of C(graphite).
ΔH = H(C(diamond)) - H(C(graphite))
ΔH = -395.4 kJ/mol - (-393.5 kJ/mol)
ΔH = -395.4 kJ/mol + 393.5 kJ/mol
ΔH = -1.9 kJ/mol
Therefore, the enthalpy of conversion of C(graphite) to C(diamond) is -1.9 kJ/mol.
Explanation:
The negative sign indicates that the conversion of C(graphite) to C(diamond) is an exothermic reaction, meaning it releases heat energy. In this case, the enthalpy change is -1.9 kJ/mol, indicating that the reaction releases 1.9 kJ of heat energy per mole of carbon converted.
Option C, which states the enthalpy of conversion as 1.9 kJ/mol, is the correct answer.
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
The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 and -395.4 KJ mol1respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol1isa)-1.9b)-788.9c)1.9d)788.9Correct answer is option 'C'. Can you explain this answer?
Question Description
The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 and -395.4 KJ mol1respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol1isa)-1.9b)-788.9c)1.9d)788.9Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 and -395.4 KJ mol1respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol1isa)-1.9b)-788.9c)1.9d)788.9Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 and -395.4 KJ mol1respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol1isa)-1.9b)-788.9c)1.9d)788.9Correct answer is option 'C'. Can you explain this answer?.
The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 and -395.4 KJ mol1respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol1isa)-1.9b)-788.9c)1.9d)788.9Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 and -395.4 KJ mol1respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol1isa)-1.9b)-788.9c)1.9d)788.9Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 and -395.4 KJ mol1respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol1isa)-1.9b)-788.9c)1.9d)788.9Correct answer is option 'C'. Can you explain this answer?.
Solutions for The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 and -395.4 KJ mol1respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol1isa)-1.9b)-788.9c)1.9d)788.9Correct answer is option 'C'. Can you explain this answer? 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 The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 and -395.4 KJ mol1respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol1isa)-1.9b)-788.9c)1.9d)788.9Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 and -395.4 KJ mol1respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol1isa)-1.9b)-788.9c)1.9d)788.9Correct answer is option 'C'. Can you explain this answer?, a detailed solution for The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 and -395.4 KJ mol1respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol1isa)-1.9b)-788.9c)1.9d)788.9Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 and -395.4 KJ mol1respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol1isa)-1.9b)-788.9c)1.9d)788.9Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The enthalpy of combustion of C(graphite) and C (diamond) are -393.5 and -395.4 KJ mol1respectively .The enthalpy of conversion of C(graphite) to C (diamond) in KJ mol1isa)-1.9b)-788.9c)1.9d)788.9Correct answer is option 'C'. Can you explain this answer? 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