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Given
C(graphite) + O2(g) →CO2(g);
ΔrHº = –393.5 kJ mol–1
H2(g) + 1/2 O2 (g) → H2O(l);
ΔrH0 = -285.8 kJ mol-1
CO2(g) + 2H2O(l) → CH4(g) + 2O2(g);
ΔrH0 = +890.3 kJ mol-1
Based on the above thermochemical equations, the value of ΔrHº at 298 K for the reaction
C(graphite) + 2H2(g) → CH4(g) wil1 be
  • a)
    +74.8 kJ mol–1
  • b)
    +144.0 kJ mol–1
  • c)
    –74.8 kJ mol–1
  • d)
    –144.0 kJ mol–1
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
GivenC(graphite) + O2(g) →CO2(g);ΔrHº = –393.5 k...


By applying the operation
(i) + 2 × (ii) + (iii), we get
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Most Upvoted Answer
GivenC(graphite) + O2(g) →CO2(g);ΔrHº = –393.5 k...
The given chemical equation is:

C(graphite) + O2(g) ->

This equation is incomplete as it does not specify the products formed from the reaction between graphite (C) and oxygen gas (O2). In order to complete the equation, the products of the reaction need to be included.
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GivenC(graphite) + O2(g) →CO2(g);ΔrHº = –393.5 kJ mol–1H2(g) + 1/2O2 (g) → H2O(l);ΔrH0 = -285.8 kJ mol-1CO2(g) + 2H2O(l) → CH4(g) + 2O2(g);ΔrH0 = +890.3 kJ mol-1Based on the above thermochemical equations, thevalue of ΔrHº at 298 K for the reactionC(graphite) + 2H2(g) → CH4(g) wil1 bea)+74.8 kJ mol–1b)+144.0 kJ mol–1c)–74.8 kJ mol–1d)–144.0 kJ mol–1Correct answer is option 'C'. Can you explain this answer?
Question Description
GivenC(graphite) + O2(g) →CO2(g);ΔrHº = –393.5 kJ mol–1H2(g) + 1/2O2 (g) → H2O(l);ΔrH0 = -285.8 kJ mol-1CO2(g) + 2H2O(l) → CH4(g) + 2O2(g);ΔrH0 = +890.3 kJ mol-1Based on the above thermochemical equations, thevalue of ΔrHº at 298 K for the reactionC(graphite) + 2H2(g) → CH4(g) wil1 bea)+74.8 kJ mol–1b)+144.0 kJ mol–1c)–74.8 kJ mol–1d)–144.0 kJ mol–1Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about GivenC(graphite) + O2(g) →CO2(g);ΔrHº = –393.5 kJ mol–1H2(g) + 1/2O2 (g) → H2O(l);ΔrH0 = -285.8 kJ mol-1CO2(g) + 2H2O(l) → CH4(g) + 2O2(g);ΔrH0 = +890.3 kJ mol-1Based on the above thermochemical equations, thevalue of ΔrHº at 298 K for the reactionC(graphite) + 2H2(g) → CH4(g) wil1 bea)+74.8 kJ mol–1b)+144.0 kJ mol–1c)–74.8 kJ mol–1d)–144.0 kJ mol–1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for GivenC(graphite) + O2(g) →CO2(g);ΔrHº = –393.5 kJ mol–1H2(g) + 1/2O2 (g) → H2O(l);ΔrH0 = -285.8 kJ mol-1CO2(g) + 2H2O(l) → CH4(g) + 2O2(g);ΔrH0 = +890.3 kJ mol-1Based on the above thermochemical equations, thevalue of ΔrHº at 298 K for the reactionC(graphite) + 2H2(g) → CH4(g) wil1 bea)+74.8 kJ mol–1b)+144.0 kJ mol–1c)–74.8 kJ mol–1d)–144.0 kJ mol–1Correct answer is option 'C'. Can you explain this answer?.
Solutions for GivenC(graphite) + O2(g) →CO2(g);ΔrHº = –393.5 kJ mol–1H2(g) + 1/2O2 (g) → H2O(l);ΔrH0 = -285.8 kJ mol-1CO2(g) + 2H2O(l) → CH4(g) + 2O2(g);ΔrH0 = +890.3 kJ mol-1Based on the above thermochemical equations, thevalue of ΔrHº at 298 K for the reactionC(graphite) + 2H2(g) → CH4(g) wil1 bea)+74.8 kJ mol–1b)+144.0 kJ mol–1c)–74.8 kJ mol–1d)–144.0 kJ mol–1Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of GivenC(graphite) + O2(g) →CO2(g);ΔrHº = –393.5 kJ mol–1H2(g) + 1/2O2 (g) → H2O(l);ΔrH0 = -285.8 kJ mol-1CO2(g) + 2H2O(l) → CH4(g) + 2O2(g);ΔrH0 = +890.3 kJ mol-1Based on the above thermochemical equations, thevalue of ΔrHº at 298 K for the reactionC(graphite) + 2H2(g) → CH4(g) wil1 bea)+74.8 kJ mol–1b)+144.0 kJ mol–1c)–74.8 kJ mol–1d)–144.0 kJ mol–1Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of GivenC(graphite) + O2(g) →CO2(g);ΔrHº = –393.5 kJ mol–1H2(g) + 1/2O2 (g) → H2O(l);ΔrH0 = -285.8 kJ mol-1CO2(g) + 2H2O(l) → CH4(g) + 2O2(g);ΔrH0 = +890.3 kJ mol-1Based on the above thermochemical equations, thevalue of ΔrHº at 298 K for the reactionC(graphite) + 2H2(g) → CH4(g) wil1 bea)+74.8 kJ mol–1b)+144.0 kJ mol–1c)–74.8 kJ mol–1d)–144.0 kJ mol–1Correct answer is option 'C'. Can you explain this answer?, a detailed solution for GivenC(graphite) + O2(g) →CO2(g);ΔrHº = –393.5 kJ mol–1H2(g) + 1/2O2 (g) → H2O(l);ΔrH0 = -285.8 kJ mol-1CO2(g) + 2H2O(l) → CH4(g) + 2O2(g);ΔrH0 = +890.3 kJ mol-1Based on the above thermochemical equations, thevalue of ΔrHº at 298 K for the reactionC(graphite) + 2H2(g) → CH4(g) wil1 bea)+74.8 kJ mol–1b)+144.0 kJ mol–1c)–74.8 kJ mol–1d)–144.0 kJ mol–1Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of GivenC(graphite) + O2(g) →CO2(g);ΔrHº = –393.5 kJ mol–1H2(g) + 1/2O2 (g) → H2O(l);ΔrH0 = -285.8 kJ mol-1CO2(g) + 2H2O(l) → CH4(g) + 2O2(g);ΔrH0 = +890.3 kJ mol-1Based on the above thermochemical equations, thevalue of ΔrHº at 298 K for the reactionC(graphite) + 2H2(g) → CH4(g) wil1 bea)+74.8 kJ mol–1b)+144.0 kJ mol–1c)–74.8 kJ mol–1d)–144.0 kJ mol–1Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice GivenC(graphite) + O2(g) →CO2(g);ΔrHº = –393.5 kJ mol–1H2(g) + 1/2O2 (g) → H2O(l);ΔrH0 = -285.8 kJ mol-1CO2(g) + 2H2O(l) → CH4(g) + 2O2(g);ΔrH0 = +890.3 kJ mol-1Based on the above thermochemical equations, thevalue of ΔrHº at 298 K for the reactionC(graphite) + 2H2(g) → CH4(g) wil1 bea)+74.8 kJ mol–1b)+144.0 kJ mol–1c)–74.8 kJ mol–1d)–144.0 kJ mol–1Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
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