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If enthalpies of formation C2H4 (g), CO2 (g) and H2O (l) at 2500C and 1 atm. pressure be 52, –394 and –286 KJ mol–1 respectively. The enthalpy of combustion of C2H4 (g) will be:
- a)+ 1412 KJ mol–1
- b)– 1412 KJ mol–1
- c)+ 141.2 KJ mol–1
- d)– 141.2 KJ mol–1
Correct answer is 'B'. Can you explain this answer?
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If enthalpies of formation C2H4 (g), CO2 (g) and H2O (l) at 2500C and ...
C2H4 + 202→2 C02 + 2 H20
Given ΔH f^0 C2H4 = 52 KJ/mole
ΔH f^0 C02 = - 394 KJ/mole
ΔHf^0 H20 = -286 KJ/mole
Most Upvoted Answer
If enthalpies of formation C2H4 (g), CO2 (g) and H2O (l) at 2500C and ...
Enthalpy of Formation:
The enthalpy of formation is the change in enthalpy that occurs when one mole of a compound is formed from its constituent elements in their standard states. It is denoted by ΔHf.
Given Enthalpies of Formation:
ΔHf(C2H4) = 52 kJ/mol
ΔHf(CO2) = 394 kJ/mol
ΔHf(H2O) = 286 kJ/mol
Enthalpy of Combustion:
The enthalpy of combustion is the change in enthalpy that occurs when one mole of a substance is completely burned in excess oxygen. It is denoted by ΔHc.
The enthalpy of combustion of C2H4 can be calculated using the enthalpies of formation of the reactants and products involved in the combustion reaction.
Combustion Reaction of C2H4:
C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (l)
To calculate the enthalpy of combustion, we need to consider the enthalpies of formation of the reactants and products in the balanced equation.
Calculating Enthalpy of Combustion:
ΔHc = [2ΔHf(CO2) + 2ΔHf(H2O)] - ΔHf(C2H4)
Substituting the given values:
ΔHc = [2(394 kJ/mol) + 2(286 kJ/mol)] - 52 kJ/mol
ΔHc = 788 kJ/mol + 572 kJ/mol - 52 kJ/mol
ΔHc = 1308 kJ/mol
Therefore, the enthalpy of combustion of C2H4 is 1308 kJ/mol.
Answer:
The correct answer is not listed among the options provided.
The enthalpy of formation is the change in enthalpy that occurs when one mole of a compound is formed from its constituent elements in their standard states. It is denoted by ΔHf.
Given Enthalpies of Formation:
ΔHf(C2H4) = 52 kJ/mol
ΔHf(CO2) = 394 kJ/mol
ΔHf(H2O) = 286 kJ/mol
Enthalpy of Combustion:
The enthalpy of combustion is the change in enthalpy that occurs when one mole of a substance is completely burned in excess oxygen. It is denoted by ΔHc.
The enthalpy of combustion of C2H4 can be calculated using the enthalpies of formation of the reactants and products involved in the combustion reaction.
Combustion Reaction of C2H4:
C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (l)
To calculate the enthalpy of combustion, we need to consider the enthalpies of formation of the reactants and products in the balanced equation.
Calculating Enthalpy of Combustion:
ΔHc = [2ΔHf(CO2) + 2ΔHf(H2O)] - ΔHf(C2H4)
Substituting the given values:
ΔHc = [2(394 kJ/mol) + 2(286 kJ/mol)] - 52 kJ/mol
ΔHc = 788 kJ/mol + 572 kJ/mol - 52 kJ/mol
ΔHc = 1308 kJ/mol
Therefore, the enthalpy of combustion of C2H4 is 1308 kJ/mol.
Answer:
The correct answer is not listed among the options provided.
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If enthalpies of formation C2H4 (g), CO2 (g) and H2O (l) at 2500C and 1 atm. pressure be 52, –394 and –286 KJ mol–1 respectively. The enthalpy of combustion of C2H4 (g) will be:a)+ 1412 KJ mol–1b)– 1412 KJ mol–1c)+ 141.2 KJ mol–1d)– 141.2 KJ mol–1Correct answer is 'B'. Can you explain this answer?
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If enthalpies of formation C2H4 (g), CO2 (g) and H2O (l) at 2500C and 1 atm. pressure be 52, –394 and –286 KJ mol–1 respectively. The enthalpy of combustion of C2H4 (g) will be:a)+ 1412 KJ mol–1b)– 1412 KJ mol–1c)+ 141.2 KJ mol–1d)– 141.2 KJ mol–1Correct answer is 'B'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about If enthalpies of formation C2H4 (g), CO2 (g) and H2O (l) at 2500C and 1 atm. pressure be 52, –394 and –286 KJ mol–1 respectively. The enthalpy of combustion of C2H4 (g) will be:a)+ 1412 KJ mol–1b)– 1412 KJ mol–1c)+ 141.2 KJ mol–1d)– 141.2 KJ mol–1Correct answer is 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If enthalpies of formation C2H4 (g), CO2 (g) and H2O (l) at 2500C and 1 atm. pressure be 52, –394 and –286 KJ mol–1 respectively. The enthalpy of combustion of C2H4 (g) will be:a)+ 1412 KJ mol–1b)– 1412 KJ mol–1c)+ 141.2 KJ mol–1d)– 141.2 KJ mol–1Correct answer is 'B'. Can you explain this answer?.
If enthalpies of formation C2H4 (g), CO2 (g) and H2O (l) at 2500C and 1 atm. pressure be 52, –394 and –286 KJ mol–1 respectively. The enthalpy of combustion of C2H4 (g) will be:a)+ 1412 KJ mol–1b)– 1412 KJ mol–1c)+ 141.2 KJ mol–1d)– 141.2 KJ mol–1Correct answer is 'B'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about If enthalpies of formation C2H4 (g), CO2 (g) and H2O (l) at 2500C and 1 atm. pressure be 52, –394 and –286 KJ mol–1 respectively. The enthalpy of combustion of C2H4 (g) will be:a)+ 1412 KJ mol–1b)– 1412 KJ mol–1c)+ 141.2 KJ mol–1d)– 141.2 KJ mol–1Correct answer is 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If enthalpies of formation C2H4 (g), CO2 (g) and H2O (l) at 2500C and 1 atm. pressure be 52, –394 and –286 KJ mol–1 respectively. The enthalpy of combustion of C2H4 (g) will be:a)+ 1412 KJ mol–1b)– 1412 KJ mol–1c)+ 141.2 KJ mol–1d)– 141.2 KJ mol–1Correct answer is 'B'. Can you explain this answer?.
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