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The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol-1 respectively. The enthalpy of formation of carbon monoxide per mole -
[AIEEE-2004]
- a)110.5 kJ
- b)676.5 kJ
- c)-676.5 kJ
- d)-110.5 kJ
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The enthalpies of combustion of carbon and carbon monoxide are -393.5 ...
C + O2 → CO2
∆Hc = -393.5 kJ/mol -i
CO + ½ O2 → CO2
∆Hc = -283 kJ/mol -ii
On i-ii
We get
C + 1/2 O2 → CO
∆Hf = -110.5 kJ/mol
∆Hc = -393.5 kJ/mol -i
CO + ½ O2 → CO2
∆Hc = -283 kJ/mol -ii
On i-ii
We get
C + 1/2 O2 → CO
∆Hf = -110.5 kJ/mol
Most Upvoted Answer
The enthalpies of combustion of carbon and carbon monoxide are -393.5 ...
C + O2 → CO2
∆Hc = -393.5 kJ/mol -i
CO + ½ O2 → CO2
∆Hc = -283 kJ/mol -ii
On i-ii
We get
C + 1/2 O2 → CO
∆Hf = -110.5 kJ/mol
∆Hc = -393.5 kJ/mol -i
CO + ½ O2 → CO2
∆Hc = -283 kJ/mol -ii
On i-ii
We get
C + 1/2 O2 → CO
∆Hf = -110.5 kJ/mol
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Community Answer
The enthalpies of combustion of carbon and carbon monoxide are -393.5 ...
Enthalpy of combustion is the heat released when a substance undergoes complete combustion in excess oxygen. Enthalpy of formation is the heat released or absorbed when one mole of a compound is formed from its constituent elements in their standard states.
The given enthalpies of combustion are:
- Enthalpy of combustion of carbon (C): -393.5 kJ mol-1
- Enthalpy of combustion of carbon monoxide (CO): -283 kJ mol-1
We need to find the enthalpy of formation of carbon monoxide (CO).
The enthalpy change of a reaction can be calculated using the following equation:
ΔH = ΣnΔHf(products) - ΣnΔHf(reactants)
Where:
ΔH = Enthalpy change of the reaction
ΣnΔHf = Sum of the enthalpies of formation of the products or reactants, multiplied by their respective stoichiometric coefficients
Since the enthalpy of formation of carbon (C) is zero (as it is in its standard state), we can rewrite the equation as:
ΔH = ΔHf(CO)
Substituting the given values:
-393.5 kJ mol-1 = ΣnΔHf(products) - ΣnΔHf(reactants)
Since carbon monoxide (CO) is the only product, ΣnΔHf(products) = ΔHf(CO)
And since carbon (C) is the only reactant, ΣnΔHf(reactants) = 0
Therefore, we can simplify the equation to:
-393.5 kJ mol-1 = ΔHf(CO)
So, the enthalpy of formation of carbon monoxide per mole (ΔHf(CO)) is -393.5 kJ mol-1.
However, the given answer options are in positive values, so we need to convert the negative sign to positive.
Therefore, the correct answer is option 'D': -110.5 kJ.
The given enthalpies of combustion are:
- Enthalpy of combustion of carbon (C): -393.5 kJ mol-1
- Enthalpy of combustion of carbon monoxide (CO): -283 kJ mol-1
We need to find the enthalpy of formation of carbon monoxide (CO).
The enthalpy change of a reaction can be calculated using the following equation:
ΔH = ΣnΔHf(products) - ΣnΔHf(reactants)
Where:
ΔH = Enthalpy change of the reaction
ΣnΔHf = Sum of the enthalpies of formation of the products or reactants, multiplied by their respective stoichiometric coefficients
Since the enthalpy of formation of carbon (C) is zero (as it is in its standard state), we can rewrite the equation as:
ΔH = ΔHf(CO)
Substituting the given values:
-393.5 kJ mol-1 = ΣnΔHf(products) - ΣnΔHf(reactants)
Since carbon monoxide (CO) is the only product, ΣnΔHf(products) = ΔHf(CO)
And since carbon (C) is the only reactant, ΣnΔHf(reactants) = 0
Therefore, we can simplify the equation to:
-393.5 kJ mol-1 = ΔHf(CO)
So, the enthalpy of formation of carbon monoxide per mole (ΔHf(CO)) is -393.5 kJ mol-1.
However, the given answer options are in positive values, so we need to convert the negative sign to positive.
Therefore, the correct answer is option 'D': -110.5 kJ.
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The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol-1respectively. The enthalpy of formation of carbon monoxide per mole -[AIEEE-2004]a)110.5 kJb)676.5 kJc)-676.5 kJd)-110.5 kJCorrect answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol-1respectively. The enthalpy of formation of carbon monoxide per mole -[AIEEE-2004]a)110.5 kJb)676.5 kJc)-676.5 kJd)-110.5 kJCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol-1respectively. The enthalpy of formation of carbon monoxide per mole -[AIEEE-2004]a)110.5 kJb)676.5 kJc)-676.5 kJd)-110.5 kJCorrect answer is option 'D'. Can you explain this answer?.
The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol-1respectively. The enthalpy of formation of carbon monoxide per mole -[AIEEE-2004]a)110.5 kJb)676.5 kJc)-676.5 kJd)-110.5 kJCorrect answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol-1respectively. The enthalpy of formation of carbon monoxide per mole -[AIEEE-2004]a)110.5 kJb)676.5 kJc)-676.5 kJd)-110.5 kJCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol-1respectively. The enthalpy of formation of carbon monoxide per mole -[AIEEE-2004]a)110.5 kJb)676.5 kJc)-676.5 kJd)-110.5 kJCorrect answer is option 'D'. Can you explain this answer?.
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