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The enthalpies of combustion of carbon and carbon monoxide are -390 kJ mol⁻1 and -278 kJ mol⁻1 respectively. The enthalpy of formation of carbon monoxide is
- a)668 kJ mol⁻1
- b)112 kJ mol⁻1
- c)-112 kJ mol⁻1
- d)-668 kJ mol⁻1
Correct answer is option 'C'. Can you explain this answer?
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The enthalpies of combustion of carbon and carbon monoxide are -390 kJ...
Enthalpy of combustion of carbon and carbon monoxide are given as:
ΔHcombustion(C) = -390 kJ/mol
ΔHcombustion(CO) = -278 kJ/mol
To find the enthalpy of formation of CO, we can use Hess's law which states that the change in enthalpy of a reaction is independent of the pathway between the initial and final states. In other words, the enthalpy change for a reaction can be calculated by combining the enthalpy changes of other reactions.
The enthalpy of formation of CO can be represented by the following equation:
C(s) + 1/2 O2(g) → CO(g)
To calculate the enthalpy of formation of CO, we can use the enthalpies of combustion of carbon and carbon monoxide, as well as the enthalpy of formation of O2.
ΔHformation(CO) = ΔHcombustion(C) + 1/2 ΔHformation(O2) - ΔHcombustion(CO)
Substituting the given values, we get:
ΔHformation(CO) = -390 kJ/mol + 1/2 (0) - (-278 kJ/mol)
ΔHformation(CO) = -390 kJ/mol + 278 kJ/mol
ΔHformation(CO) = -112 kJ/mol
Therefore, the correct answer is option C: -112 kJ/mol.
ΔHcombustion(C) = -390 kJ/mol
ΔHcombustion(CO) = -278 kJ/mol
To find the enthalpy of formation of CO, we can use Hess's law which states that the change in enthalpy of a reaction is independent of the pathway between the initial and final states. In other words, the enthalpy change for a reaction can be calculated by combining the enthalpy changes of other reactions.
The enthalpy of formation of CO can be represented by the following equation:
C(s) + 1/2 O2(g) → CO(g)
To calculate the enthalpy of formation of CO, we can use the enthalpies of combustion of carbon and carbon monoxide, as well as the enthalpy of formation of O2.
ΔHformation(CO) = ΔHcombustion(C) + 1/2 ΔHformation(O2) - ΔHcombustion(CO)
Substituting the given values, we get:
ΔHformation(CO) = -390 kJ/mol + 1/2 (0) - (-278 kJ/mol)
ΔHformation(CO) = -390 kJ/mol + 278 kJ/mol
ΔHformation(CO) = -112 kJ/mol
Therefore, the correct answer is option C: -112 kJ/mol.
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The enthalpies of combustion of carbon and carbon monoxide are -390 kJ mol⁻1 and -278 kJ mol⁻1 respectively. The enthalpy of formation of carbon monoxide isa)668 kJ mol⁻1b)112 kJ mol⁻1c)-112 kJ mol⁻1d)-668 kJ mol⁻1Correct answer is option 'C'. Can you explain this answer?
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The enthalpies of combustion of carbon and carbon monoxide are -390 kJ mol⁻1 and -278 kJ mol⁻1 respectively. The enthalpy of formation of carbon monoxide isa)668 kJ mol⁻1b)112 kJ mol⁻1c)-112 kJ mol⁻1d)-668 kJ mol⁻1Correct 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 enthalpies of combustion of carbon and carbon monoxide are -390 kJ mol⁻1 and -278 kJ mol⁻1 respectively. The enthalpy of formation of carbon monoxide isa)668 kJ mol⁻1b)112 kJ mol⁻1c)-112 kJ mol⁻1d)-668 kJ mol⁻1Correct 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 enthalpies of combustion of carbon and carbon monoxide are -390 kJ mol⁻1 and -278 kJ mol⁻1 respectively. The enthalpy of formation of carbon monoxide isa)668 kJ mol⁻1b)112 kJ mol⁻1c)-112 kJ mol⁻1d)-668 kJ mol⁻1Correct answer is option 'C'. Can you explain this answer?.
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Solutions for The enthalpies of combustion of carbon and carbon monoxide are -390 kJ mol⁻1 and -278 kJ mol⁻1 respectively. The enthalpy of formation of carbon monoxide isa)668 kJ mol⁻1b)112 kJ mol⁻1c)-112 kJ mol⁻1d)-668 kJ mol⁻1Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
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