Class 11 Exam > Class 11 Questions > The value of enthalpy change (DH) for the rea...
Start Learning for Free
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?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
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.
Most Upvoted Answer
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.
Free Test
FREE
| Start Free Test |
Community Answer
The value of enthalpy change (DH) for the reactionC2H5OH(l)+ 3O2(g)&ra...
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
|
Explore Courses for Class 11 exam
|
|
Top Courses for Class 11View all
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?
Question Description
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?.
Solutions 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? in English & in Hindi are available as part of our courses for Class 11.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of
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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup