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The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are - 890.3 kJ mol-1, - 393.5 kJ mol-1 and - 285.8 kJ mol-1 respectively. Enthalpy of formation of CH4(g) will be
- a)-74.8 kJ mol-1
- b)-52.57 kJ mol-1
- c)+74.8 kJ mol-1
- d)+52.27 kJ mol-1
Correct answer is option 'A'. Can you explain this answer?
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Enthalpy of Combustion and Enthalpy of Formation
Enthalpy of combustion (ΔHcomb) is the amount of heat released when one mole of a substance undergoes complete combustion in excess oxygen. It is a measure of the energy content of the substance.
Enthalpy of formation (ΔHf) is the amount of heat absorbed or released when one mole of a compound is formed from its constituent elements in their standard states. It is a measure of the stability of the compound.
Given Data:
ΔHcomb of methane (CH4) = -890.3 kJ mol-1
ΔHcomb of graphite (C) = -393.5 kJ mol-1
ΔHcomb of dihydrogen (H2) = -285.8 kJ mol-1
Using these values, we can calculate the enthalpy of formation of methane (CH4).
Equation 1: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Equation 2: C(s) + O2(g) → CO2(g)
Equation 3: H2(g) + 1/2O2(g) → H2O(l)
We can use these equations to manipulate the given enthalpy values and obtain the enthalpy of formation of methane.
Calculations:
1. Write the balanced equation for the combustion of methane:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
2. Manipulate the equation to obtain the enthalpy of formation of methane:
ΔHcomb(CH4) = ΔHf(CO2) + 2ΔHf(H2O) - ΔHf(CH4)
3. Substitute the given enthalpy values:
-890.3 kJ mol-1 = ΔHf(CO2) + 2(-285.8 kJ mol-1) - ΔHf(CH4)
4. Substitute the enthalpy of formation of CO2 and H2O:
-890.3 kJ mol-1 = -393.5 kJ mol-1 + 2(-285.8 kJ mol-1) - ΔHf(CH4)
5. Simplify the equation:
-890.3 kJ mol-1 = -965.1 kJ mol-1 - ΔHf(CH4)
6. Solve for ΔHf(CH4):
ΔHf(CH4) = -965.1 kJ mol-1 + 890.3 kJ mol-1
ΔHf(CH4) = -74.8 kJ mol-1
Therefore, the enthalpy of formation of methane (CH4) is -74.8 kJ mol-1.
Enthalpy of combustion (ΔHcomb) is the amount of heat released when one mole of a substance undergoes complete combustion in excess oxygen. It is a measure of the energy content of the substance.
Enthalpy of formation (ΔHf) is the amount of heat absorbed or released when one mole of a compound is formed from its constituent elements in their standard states. It is a measure of the stability of the compound.
Given Data:
ΔHcomb of methane (CH4) = -890.3 kJ mol-1
ΔHcomb of graphite (C) = -393.5 kJ mol-1
ΔHcomb of dihydrogen (H2) = -285.8 kJ mol-1
Using these values, we can calculate the enthalpy of formation of methane (CH4).
Equation 1: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Equation 2: C(s) + O2(g) → CO2(g)
Equation 3: H2(g) + 1/2O2(g) → H2O(l)
We can use these equations to manipulate the given enthalpy values and obtain the enthalpy of formation of methane.
Calculations:
1. Write the balanced equation for the combustion of methane:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
2. Manipulate the equation to obtain the enthalpy of formation of methane:
ΔHcomb(CH4) = ΔHf(CO2) + 2ΔHf(H2O) - ΔHf(CH4)
3. Substitute the given enthalpy values:
-890.3 kJ mol-1 = ΔHf(CO2) + 2(-285.8 kJ mol-1) - ΔHf(CH4)
4. Substitute the enthalpy of formation of CO2 and H2O:
-890.3 kJ mol-1 = -393.5 kJ mol-1 + 2(-285.8 kJ mol-1) - ΔHf(CH4)
5. Simplify the equation:
-890.3 kJ mol-1 = -965.1 kJ mol-1 - ΔHf(CH4)
6. Solve for ΔHf(CH4):
ΔHf(CH4) = -965.1 kJ mol-1 + 890.3 kJ mol-1
ΔHf(CH4) = -74.8 kJ mol-1
Therefore, the enthalpy of formation of methane (CH4) is -74.8 kJ mol-1.
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The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are - 890.3 kJ mol-1, - 393.5 kJ mol-1 and - 285.8 kJ mol-1 respectively. Enthalpy of formation of CH4(g) will bea)-74.8 kJ mol-1b)-52.57kJ mol-1c)+74.8kJ mol-1d)+52.27kJ mol-1Correct answer is option 'A'. Can you explain this answer?
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The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are - 890.3 kJ mol-1, - 393.5 kJ mol-1 and - 285.8 kJ mol-1 respectively. Enthalpy of formation of CH4(g) will bea)-74.8 kJ mol-1b)-52.57kJ mol-1c)+74.8kJ mol-1d)+52.27kJ mol-1Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are - 890.3 kJ mol-1, - 393.5 kJ mol-1 and - 285.8 kJ mol-1 respectively. Enthalpy of formation of CH4(g) will bea)-74.8 kJ mol-1b)-52.57kJ mol-1c)+74.8kJ mol-1d)+52.27kJ mol-1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are - 890.3 kJ mol-1, - 393.5 kJ mol-1 and - 285.8 kJ mol-1 respectively. Enthalpy of formation of CH4(g) will bea)-74.8 kJ mol-1b)-52.57kJ mol-1c)+74.8kJ mol-1d)+52.27kJ mol-1Correct answer is option 'A'. Can you explain this answer?.
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are - 890.3 kJ mol-1, - 393.5 kJ mol-1 and - 285.8 kJ mol-1 respectively. Enthalpy of formation of CH4(g) will bea)-74.8 kJ mol-1b)-52.57kJ mol-1c)+74.8kJ mol-1d)+52.27kJ mol-1Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are - 890.3 kJ mol-1, - 393.5 kJ mol-1 and - 285.8 kJ mol-1 respectively. Enthalpy of formation of CH4(g) will bea)-74.8 kJ mol-1b)-52.57kJ mol-1c)+74.8kJ mol-1d)+52.27kJ mol-1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are - 890.3 kJ mol-1, - 393.5 kJ mol-1 and - 285.8 kJ mol-1 respectively. Enthalpy of formation of CH4(g) will bea)-74.8 kJ mol-1b)-52.57kJ mol-1c)+74.8kJ mol-1d)+52.27kJ mol-1Correct answer is option 'A'. Can you explain this answer?.
Solutions for The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are - 890.3 kJ mol-1, - 393.5 kJ mol-1 and - 285.8 kJ mol-1 respectively. Enthalpy of formation of CH4(g) will bea)-74.8 kJ mol-1b)-52.57kJ mol-1c)+74.8kJ mol-1d)+52.27kJ mol-1Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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