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Given the following thermochemical data at 298 K and 1 bar
ΔH°vap (CH3OH) = 38.0 kJ mol-1
ΔfH°: H(g) = 218 kJ mol-1
O(g) = 249 kJ mol-1
C(g) = 715 kJ mol-1
ΔfH°: H(g) = 218 kJ mol-1
O(g) = 249 kJ mol-1
C(g) = 715 kJ mol-1
Bond dissociation energy
(C—H) = 415 kJ mol-1
(C—O) = 356 kJ mol-1
(O—H) = 4 6 3 kJ mol-1
(C—O) = 356 kJ mol-1
(O—H) = 4 6 3 kJ mol-1
Q. The ΔfH° of liquid methyl alcohol in kJ mol-1 is
- a)-170 kJ mol-1
- b)+170 kJ mol-1
- c)-266 kJ mol-1
- d)+266 kJ mol-1
Correct answer is option 'D'. Can you explain this answer?
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Given the following thermochemical data at 298 K and 1 barΔH°...
Thermochemical Data
Given thermochemical data at 298 K and 1 bar:
- Hvap (CH3OH) = 38.0 kJ mol-1
- fH: H(g) = 218 kJ mol-1
- O(g) = 249 kJ mol-1
- C(g) = 715 kJ mol-1
- Bond dissociation energy:
- (CH) = 415 kJ mol-1
- (CO) = 356 kJ mol-1
- (OH) = 463 kJ mol-1
Finding the fH of Liquid Methyl Alcohol
To find the fH of liquid methyl alcohol (CH3OH), we need to consider the enthalpy changes involved in the formation of liquid methyl alcohol from its constituent elements.
The formation of liquid methyl alcohol can be represented by the following equation:
CH3OH(l) = C(g) + 2H(g) + O(g)
The enthalpy change of this reaction can be calculated by summing the enthalpies of formation of the products and subtracting the enthalpies of formation of the reactants.
The enthalpy change for the formation of liquid methyl alcohol can be expressed as:
ΔHf(CH3OH) = fH(CH3OH) - [fH(C(g)) + 2fH(H(g)) + fH(O(g))]
Substituting the given values:
ΔHf(CH3OH) = fH(CH3OH) - [715 kJ mol-1 + 2(218 kJ mol-1) + 249 kJ mol-1]
Simplifying:
ΔHf(CH3OH) = fH(CH3OH) - 1400 kJ mol-1
We are given that the enthalpy of vaporization (Hvap) of CH3OH is 38.0 kJ mol-1. The enthalpy of vaporization can be related to the enthalpy of formation as follows:
Hvap(CH3OH) = ΔHf(CH3OH) - ΔHf(g)
Where ΔHf(g) is the enthalpy of formation of gaseous CH3OH. Since we are given the enthalpy of formation of H(g), O(g), and C(g), we can calculate ΔHf(g) as follows:
ΔHf(g) = fH(CH3OH) - [2fH(H(g)) + fH(O(g)) + fH(C(g))]
Substituting the given values:
38.0 kJ mol-1 = ΔHf(CH3OH) - [2(218 kJ mol-1) + 249 kJ mol-1 + 715 kJ mol-1]
Simplifying:
38.0 kJ mol-1 = ΔHf(CH3OH) - 1400 kJ mol-1
Rearranging the equation:
ΔHf(CH3OH) = 1438 kJ mol-1
Therefore, the
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Given the following thermochemical data at 298 K and 1 barΔH°vap (CH3OH) = 38.0 kJ mol-1ΔfH°: H(g) = 218 kJ mol-1O(g) = 249 kJ mol-1C(g) = 715 kJ mol-1Bond dissociation energy(C—H) = 415 kJ mol-1(C—O) = 356 kJ mol-1(O—H) = 4 6 3 kJ mol-1Q.The ΔfH° of liquid methyl alcohol in kJ mol-1 isa)-170 kJ mol-1b)+170 kJ mol-1c)-266 kJ mol-1d)+266 kJ mol-1Correct answer is option 'D'. Can you explain this answer?
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Given the following thermochemical data at 298 K and 1 barΔH°vap (CH3OH) = 38.0 kJ mol-1ΔfH°: H(g) = 218 kJ mol-1O(g) = 249 kJ mol-1C(g) = 715 kJ mol-1Bond dissociation energy(C—H) = 415 kJ mol-1(C—O) = 356 kJ mol-1(O—H) = 4 6 3 kJ mol-1Q.The ΔfH° of liquid methyl alcohol in kJ mol-1 isa)-170 kJ mol-1b)+170 kJ mol-1c)-266 kJ mol-1d)+266 kJ mol-1Correct answer is option 'D'. 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 Given the following thermochemical data at 298 K and 1 barΔH°vap (CH3OH) = 38.0 kJ mol-1ΔfH°: H(g) = 218 kJ mol-1O(g) = 249 kJ mol-1C(g) = 715 kJ mol-1Bond dissociation energy(C—H) = 415 kJ mol-1(C—O) = 356 kJ mol-1(O—H) = 4 6 3 kJ mol-1Q.The ΔfH° of liquid methyl alcohol in kJ mol-1 isa)-170 kJ mol-1b)+170 kJ mol-1c)-266 kJ mol-1d)+266 kJ mol-1Correct answer is option 'D'. 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 Given the following thermochemical data at 298 K and 1 barΔH°vap (CH3OH) = 38.0 kJ mol-1ΔfH°: H(g) = 218 kJ mol-1O(g) = 249 kJ mol-1C(g) = 715 kJ mol-1Bond dissociation energy(C—H) = 415 kJ mol-1(C—O) = 356 kJ mol-1(O—H) = 4 6 3 kJ mol-1Q.The ΔfH° of liquid methyl alcohol in kJ mol-1 isa)-170 kJ mol-1b)+170 kJ mol-1c)-266 kJ mol-1d)+266 kJ mol-1Correct answer is option 'D'. Can you explain this answer?.
Given the following thermochemical data at 298 K and 1 barΔH°vap (CH3OH) = 38.0 kJ mol-1ΔfH°: H(g) = 218 kJ mol-1O(g) = 249 kJ mol-1C(g) = 715 kJ mol-1Bond dissociation energy(C—H) = 415 kJ mol-1(C—O) = 356 kJ mol-1(O—H) = 4 6 3 kJ mol-1Q.The ΔfH° of liquid methyl alcohol in kJ mol-1 isa)-170 kJ mol-1b)+170 kJ mol-1c)-266 kJ mol-1d)+266 kJ mol-1Correct answer is option 'D'. 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 Given the following thermochemical data at 298 K and 1 barΔH°vap (CH3OH) = 38.0 kJ mol-1ΔfH°: H(g) = 218 kJ mol-1O(g) = 249 kJ mol-1C(g) = 715 kJ mol-1Bond dissociation energy(C—H) = 415 kJ mol-1(C—O) = 356 kJ mol-1(O—H) = 4 6 3 kJ mol-1Q.The ΔfH° of liquid methyl alcohol in kJ mol-1 isa)-170 kJ mol-1b)+170 kJ mol-1c)-266 kJ mol-1d)+266 kJ mol-1Correct answer is option 'D'. 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 Given the following thermochemical data at 298 K and 1 barΔH°vap (CH3OH) = 38.0 kJ mol-1ΔfH°: H(g) = 218 kJ mol-1O(g) = 249 kJ mol-1C(g) = 715 kJ mol-1Bond dissociation energy(C—H) = 415 kJ mol-1(C—O) = 356 kJ mol-1(O—H) = 4 6 3 kJ mol-1Q.The ΔfH° of liquid methyl alcohol in kJ mol-1 isa)-170 kJ mol-1b)+170 kJ mol-1c)-266 kJ mol-1d)+266 kJ mol-1Correct answer is option 'D'. Can you explain this answer?.
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