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The value of enthalpy change (DH) for the reaction C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
at 27°C is -1366.5 kJ mol-1. The value of internal energy change for the above reaction at this temperature will be -
[AIEEE 2011]
- a)-1369.0 kJ
- b)-1364.0 kJ
- c)-1361.5 kJ
- d)-1371.5 kJ
Correct answer is option 'B'. Can you explain this answer?
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The value of enthalpy change (DH) for the reactionC2H5OH(l)+ 3O2(g)&ra...
The correct answer is option B
The change in the number of moles of gaseous species Δn = 2−3 =−1
ΔH = ΔE+ΔnRT
−1366.5 kJ/mol = ΔE+[(−1)×8.314×10−3 kJ/mol/K×(27+273) K]
ΔE = −1364.0 kJ/mol
The value of internal energy change for the reaction is −1364.0 kJ/mol.
The change in the number of moles of gaseous species Δn = 2−3 =−1
ΔH = ΔE+ΔnRT
−1366.5 kJ/mol = ΔE+[(−1)×8.314×10−3 kJ/mol/K×(27+273) K]
ΔE = −1364.0 kJ/mol
The value of internal energy change for the reaction is −1364.0 kJ/mol.
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The value of enthalpy change (DH) for the reactionC2H5OH(l)+ 3O2(g)&ra...
The value of enthalpy change (ΔH) for the reaction C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) can be determined by looking up the standard enthalpies of formation for each compound involved in the reaction and using Hess's Law.
The standard enthalpy of formation (ΔHf°) is the enthalpy change that occurs when one mole of a compound is formed from its elements in their standard states at a given temperature and pressure.
Using the standard enthalpies of formation values:
ΔHf° (C2H5OH(l)) = -277.6 kJ/mol
ΔHf° (O2(g)) = 0 kJ/mol
ΔHf° (CO2(g)) = -393.5 kJ/mol
ΔHf° (H2O(l)) = -285.8 kJ/mol
According to Hess's Law, the enthalpy change of a reaction can be calculated by taking the difference between the sum of the standard enthalpies of formation of the products and the sum of the standard enthalpies of formation of the reactants, multiplied by their stoichiometric coefficients.
ΔH = [2ΔHf° (CO2(g)) + 3ΔHf° (H2O(l))] - [ΔHf° (C2H5OH(l)) + 3ΔHf° (O2(g))]
ΔH = [2(-393.5 kJ/mol) + 3(-285.8 kJ/mol)] - [(-277.6 kJ/mol) + 3(0 kJ/mol)]
ΔH = [-787.0 kJ/mol + (-857.4 kJ/mol)] - [-277.6 kJ/mol]
ΔH = -1644.4 kJ/mol + 277.6 kJ/mol
ΔH = -1366.8 kJ/mol
Therefore, the value of enthalpy change (ΔH) for the reaction C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) is -1366.8 kJ/mol.
The standard enthalpy of formation (ΔHf°) is the enthalpy change that occurs when one mole of a compound is formed from its elements in their standard states at a given temperature and pressure.
Using the standard enthalpies of formation values:
ΔHf° (C2H5OH(l)) = -277.6 kJ/mol
ΔHf° (O2(g)) = 0 kJ/mol
ΔHf° (CO2(g)) = -393.5 kJ/mol
ΔHf° (H2O(l)) = -285.8 kJ/mol
According to Hess's Law, the enthalpy change of a reaction can be calculated by taking the difference between the sum of the standard enthalpies of formation of the products and the sum of the standard enthalpies of formation of the reactants, multiplied by their stoichiometric coefficients.
ΔH = [2ΔHf° (CO2(g)) + 3ΔHf° (H2O(l))] - [ΔHf° (C2H5OH(l)) + 3ΔHf° (O2(g))]
ΔH = [2(-393.5 kJ/mol) + 3(-285.8 kJ/mol)] - [(-277.6 kJ/mol) + 3(0 kJ/mol)]
ΔH = [-787.0 kJ/mol + (-857.4 kJ/mol)] - [-277.6 kJ/mol]
ΔH = -1644.4 kJ/mol + 277.6 kJ/mol
ΔH = -1366.8 kJ/mol
Therefore, the value of enthalpy change (ΔH) for the reaction C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) is -1366.8 kJ/mol.
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The value of enthalpy change (DH) for the reactionC2H5OH(l)+ 3O2(g)→2CO2(g)+ 3H2O(l)at 27°C is -1366.5 kJ mol-1. The value of internal energy change for the above reaction at this temperature will be - [AIEEE 2011]a)-1369.0 kJb)-1364.0 kJc)-1361.5 kJd)-1371.5 kJCorrect answer is option 'B'. Can you explain this answer?
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The value of enthalpy change (DH) for the reactionC2H5OH(l)+ 3O2(g)→2CO2(g)+ 3H2O(l)at 27°C is -1366.5 kJ mol-1. The value of internal energy change for the above reaction at this temperature will be - [AIEEE 2011]a)-1369.0 kJb)-1364.0 kJc)-1361.5 kJd)-1371.5 kJCorrect answer is option 'B'. 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 value of enthalpy change (DH) for the reactionC2H5OH(l)+ 3O2(g)→2CO2(g)+ 3H2O(l)at 27°C is -1366.5 kJ mol-1. The value of internal energy change for the above reaction at this temperature will be - [AIEEE 2011]a)-1369.0 kJb)-1364.0 kJc)-1361.5 kJd)-1371.5 kJCorrect answer is option 'B'. 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 value of enthalpy change (DH) for the reactionC2H5OH(l)+ 3O2(g)→2CO2(g)+ 3H2O(l)at 27°C is -1366.5 kJ mol-1. The value of internal energy change for the above reaction at this temperature will be - [AIEEE 2011]a)-1369.0 kJb)-1364.0 kJc)-1361.5 kJd)-1371.5 kJCorrect answer is option 'B'. Can you explain this answer?.
The value of enthalpy change (DH) for the reactionC2H5OH(l)+ 3O2(g)→2CO2(g)+ 3H2O(l)at 27°C is -1366.5 kJ mol-1. The value of internal energy change for the above reaction at this temperature will be - [AIEEE 2011]a)-1369.0 kJb)-1364.0 kJc)-1361.5 kJd)-1371.5 kJCorrect answer is option 'B'. 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 value of enthalpy change (DH) for the reactionC2H5OH(l)+ 3O2(g)→2CO2(g)+ 3H2O(l)at 27°C is -1366.5 kJ mol-1. The value of internal energy change for the above reaction at this temperature will be - [AIEEE 2011]a)-1369.0 kJb)-1364.0 kJc)-1361.5 kJd)-1371.5 kJCorrect answer is option 'B'. 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 value of enthalpy change (DH) for the reactionC2H5OH(l)+ 3O2(g)→2CO2(g)+ 3H2O(l)at 27°C is -1366.5 kJ mol-1. The value of internal energy change for the above reaction at this temperature will be - [AIEEE 2011]a)-1369.0 kJb)-1364.0 kJc)-1361.5 kJd)-1371.5 kJCorrect answer is option 'B'. Can you explain this answer?.
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