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Calculate the heat of combustion of ethane, in the reaction C2H6(g) + 3 1/2O2(g) → 2CO2(g) + 3H2O(l) where the heats of formation of ethane gas, carbon dioxide gas and water liquid are –84.7 kJ mol-1, -393.5 kJ mol-1 and –285.8 kJ mol-1 respectively.
- a)-1559.7 kJ mol-1
- b)-155.7 kJ mol-1
- c)-159.7 kJ mol-1
- d)-559.7 kJ mol-1
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Calculate the heat of combustion of ethane, in the reaction C2H6(g) + ...
The balanced equation for the combustion of ethane is:
C2H6(g) + 3 1/2O2(g) → 2CO2(g) + 3H2O(l)
The heat of combustion (ΔHc) is the amount of heat released when one mole of a substance is burned completely in oxygen under standard conditions (25°C and 1 atm). To calculate the ΔHc of ethane, we need to use the standard enthalpies of formation (ΔHf°) of the products and reactants.
The standard enthalpy of formation is the amount of heat released or absorbed when one mole of a compound is formed from its elements in their standard states (usually at 25°C and 1 atm). The ΔHf° values for the compounds in the balanced equation are:
ΔHf°(C2H6) = -84.68 kJ/mol
ΔHf°(CO2) = -393.51 kJ/mol
ΔHf°(H2O) = -285.83 kJ/mol
ΔHf°(O2) = 0 kJ/mol
Using these values and the stoichiometry of the balanced equation, we can calculate the ΔHc of ethane:
ΔHc = Σ(nΔHf°[products]) - Σ(nΔHf°[reactants])
= [2ΔHf°(CO2) + 3ΔHf°(H2O)] - [ΔHf°(C2H6) + 3 1/2ΔHf°(O2)]
= [2(-393.51) + 3(-285.83)] - [-84.68 + 3 1/2(0)]
= -1560.96 kJ/mol
Therefore, the heat of combustion of ethane is -1560.96 kJ/mol. This means that when one mole of ethane is burned completely in oxygen, 1560.96 kJ of heat is released.
C2H6(g) + 3 1/2O2(g) → 2CO2(g) + 3H2O(l)
The heat of combustion (ΔHc) is the amount of heat released when one mole of a substance is burned completely in oxygen under standard conditions (25°C and 1 atm). To calculate the ΔHc of ethane, we need to use the standard enthalpies of formation (ΔHf°) of the products and reactants.
The standard enthalpy of formation is the amount of heat released or absorbed when one mole of a compound is formed from its elements in their standard states (usually at 25°C and 1 atm). The ΔHf° values for the compounds in the balanced equation are:
ΔHf°(C2H6) = -84.68 kJ/mol
ΔHf°(CO2) = -393.51 kJ/mol
ΔHf°(H2O) = -285.83 kJ/mol
ΔHf°(O2) = 0 kJ/mol
Using these values and the stoichiometry of the balanced equation, we can calculate the ΔHc of ethane:
ΔHc = Σ(nΔHf°[products]) - Σ(nΔHf°[reactants])
= [2ΔHf°(CO2) + 3ΔHf°(H2O)] - [ΔHf°(C2H6) + 3 1/2ΔHf°(O2)]
= [2(-393.51) + 3(-285.83)] - [-84.68 + 3 1/2(0)]
= -1560.96 kJ/mol
Therefore, the heat of combustion of ethane is -1560.96 kJ/mol. This means that when one mole of ethane is burned completely in oxygen, 1560.96 kJ of heat is released.
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Calculate the heat of combustion of ethane, in the reaction C2H6(g) + ...
Given that C2H6(g) + 3 1/2 O2(g) → 2CO2(g) + 3H2O(l) As we know that, ΔHr = Σ ΔHf[products] – Σ ΔHf[reactants]; ΔHc(C2H6) = 2 X (-393.5 kJ mol-1) + 3 X (-285.8 kJ mol-1) – (- 84.7 kJ mol-1) – 2 X (0 kJ mol-1) = ΔHc(C2H6) = – 1559.7 kJ mol-1.
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Calculate the heat of combustion of ethane, in the reaction C2H6(g) + 3 1/2O2(g) → 2CO2(g) + 3H2O(l) where the heats of formation of ethane gas, carbon dioxide gas and water liquid are –84.7 kJ mol-1, -393.5 kJ mol-1 and –285.8 kJ mol-1 respectively.a)-1559.7 kJ mol-1b)-155.7 kJ mol-1c)-159.7 kJ mol-1d)-559.7 kJ mol-1Correct answer is option 'A'. Can you explain this answer?
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Calculate the heat of combustion of ethane, in the reaction C2H6(g) + 3 1/2O2(g) → 2CO2(g) + 3H2O(l) where the heats of formation of ethane gas, carbon dioxide gas and water liquid are –84.7 kJ mol-1, -393.5 kJ mol-1 and –285.8 kJ mol-1 respectively.a)-1559.7 kJ mol-1b)-155.7 kJ mol-1c)-159.7 kJ mol-1d)-559.7 kJ mol-1Correct answer is option 'A'. 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 Calculate the heat of combustion of ethane, in the reaction C2H6(g) + 3 1/2O2(g) → 2CO2(g) + 3H2O(l) where the heats of formation of ethane gas, carbon dioxide gas and water liquid are –84.7 kJ mol-1, -393.5 kJ mol-1 and –285.8 kJ mol-1 respectively.a)-1559.7 kJ mol-1b)-155.7 kJ mol-1c)-159.7 kJ mol-1d)-559.7 kJ mol-1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the heat of combustion of ethane, in the reaction C2H6(g) + 3 1/2O2(g) → 2CO2(g) + 3H2O(l) where the heats of formation of ethane gas, carbon dioxide gas and water liquid are –84.7 kJ mol-1, -393.5 kJ mol-1 and –285.8 kJ mol-1 respectively.a)-1559.7 kJ mol-1b)-155.7 kJ mol-1c)-159.7 kJ mol-1d)-559.7 kJ mol-1Correct answer is option 'A'. Can you explain this answer?.
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