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The standard enthalpy of formation of Nh3 is 46.0 kJ per mol if the enthalpy of formation of H2 from its atom is -436 kJ per mol and that of N2 is -712 kJ per mol the average Bond enthalpy of N-H bond in Nh3 is ?? Ans is +352 kJ per mole.plz give step.?
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The standard enthalpy of formation of Nh3 is 46.0 kJ per mol if the en...
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The standard enthalpy of formation of Nh3 is 46.0 kJ per mol if the en...
Given:
Standard enthalpy of formation of NH3 = 46.0 kJ/mol
Enthalpy of formation of H2 from its atom = -436 kJ/mol
Enthalpy of formation of N2 = -712 kJ/mol
To find: Average Bond enthalpy of N-H bond in NH3
Solution:
The standard enthalpy of formation of NH3 can be written as follows:
ΔHf°(NH3) = ΣΔHf°(products) - ΣΔHf°(reactants)
Since NH3 is the only product, we can write:
ΔHf°(NH3) = ΔHf°(NH3) - [ΔHf°(N2) + 3ΔHf°(H2)]
Substituting the given values, we get:
46.0 kJ/mol = ΔHf°(NH3) - [-712 kJ/mol + 3(-436 kJ/mol)]
Simplifying, we get:
ΔHf°(NH3) = -46.0 kJ/mol + 712 kJ/mol - 3(-436 kJ/mol)
ΔHf°(NH3) = -46.0 kJ/mol + 712 kJ/mol + 1308 kJ/mol
ΔHf°(NH3) = 1974 kJ/mol
Now, we can use the concept of bond enthalpy to find the average bond enthalpy of N-H bond in NH3.
The average bond enthalpy of N-H bond is the energy required to break one mole of N-H bonds in NH3.
We know that NH3 contains three N-H bonds.
So, the average bond enthalpy of N-H bond can be calculated as:
Average bond enthalpy of N-H bond = [Σ (bond enthalpy of N-H)] / 3
We can use the given enthalpies of formation to calculate the bond enthalpy of N-H bond.
ΔHf°(NH3) = ΣΔHf°(products) - ΣΔHf°(reactants)
ΔHf°(NH3) = 3ΔH(N-H) + ΔH(N-N) + 3ΔH(H-H)
Substituting the given values, we get:
1974 kJ/mol = 3ΔH(N-H) + (-712 kJ/mol) + 3(-436 kJ/mol)
Simplifying, we get:
3ΔH(N-H) = 1974 kJ/mol + 712 kJ/mol + 3(436 kJ/mol)
3ΔH(N-H) = 3522 kJ/mol
ΔH(N-H) = 1174 kJ/mol
Finally, the average bond enthalpy of N-H bond can be calculated as:
Average bond enthalpy of N-H bond = [3ΔH(N-H)] / 3
Average bond enthalpy of N-H bond = 1174 kJ/mol / 3
Average bond enthalpy of N-H bond = 352 kJ/mol
Therefore, the average
Standard enthalpy of formation of NH3 = 46.0 kJ/mol
Enthalpy of formation of H2 from its atom = -436 kJ/mol
Enthalpy of formation of N2 = -712 kJ/mol
To find: Average Bond enthalpy of N-H bond in NH3
Solution:
The standard enthalpy of formation of NH3 can be written as follows:
ΔHf°(NH3) = ΣΔHf°(products) - ΣΔHf°(reactants)
Since NH3 is the only product, we can write:
ΔHf°(NH3) = ΔHf°(NH3) - [ΔHf°(N2) + 3ΔHf°(H2)]
Substituting the given values, we get:
46.0 kJ/mol = ΔHf°(NH3) - [-712 kJ/mol + 3(-436 kJ/mol)]
Simplifying, we get:
ΔHf°(NH3) = -46.0 kJ/mol + 712 kJ/mol - 3(-436 kJ/mol)
ΔHf°(NH3) = -46.0 kJ/mol + 712 kJ/mol + 1308 kJ/mol
ΔHf°(NH3) = 1974 kJ/mol
Now, we can use the concept of bond enthalpy to find the average bond enthalpy of N-H bond in NH3.
The average bond enthalpy of N-H bond is the energy required to break one mole of N-H bonds in NH3.
We know that NH3 contains three N-H bonds.
So, the average bond enthalpy of N-H bond can be calculated as:
Average bond enthalpy of N-H bond = [Σ (bond enthalpy of N-H)] / 3
We can use the given enthalpies of formation to calculate the bond enthalpy of N-H bond.
ΔHf°(NH3) = ΣΔHf°(products) - ΣΔHf°(reactants)
ΔHf°(NH3) = 3ΔH(N-H) + ΔH(N-N) + 3ΔH(H-H)
Substituting the given values, we get:
1974 kJ/mol = 3ΔH(N-H) + (-712 kJ/mol) + 3(-436 kJ/mol)
Simplifying, we get:
3ΔH(N-H) = 1974 kJ/mol + 712 kJ/mol + 3(436 kJ/mol)
3ΔH(N-H) = 3522 kJ/mol
ΔH(N-H) = 1174 kJ/mol
Finally, the average bond enthalpy of N-H bond can be calculated as:
Average bond enthalpy of N-H bond = [3ΔH(N-H)] / 3
Average bond enthalpy of N-H bond = 1174 kJ/mol / 3
Average bond enthalpy of N-H bond = 352 kJ/mol
Therefore, the average
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The standard enthalpy of formation of Nh3 is 46.0 kJ per mol if the enthalpy of formation of H2 from its atom is -436 kJ per mol and that of N2 is -712 kJ per mol the average Bond enthalpy of N-H bond in Nh3 is ?? Ans is +352 kJ per mole.plz give step.?
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The standard enthalpy of formation of Nh3 is 46.0 kJ per mol if the enthalpy of formation of H2 from its atom is -436 kJ per mol and that of N2 is -712 kJ per mol the average Bond enthalpy of N-H bond in Nh3 is ?? Ans is +352 kJ per mole.plz give step.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The standard enthalpy of formation of Nh3 is 46.0 kJ per mol if the enthalpy of formation of H2 from its atom is -436 kJ per mol and that of N2 is -712 kJ per mol the average Bond enthalpy of N-H bond in Nh3 is ?? Ans is +352 kJ per mole.plz give step.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The standard enthalpy of formation of Nh3 is 46.0 kJ per mol if the enthalpy of formation of H2 from its atom is -436 kJ per mol and that of N2 is -712 kJ per mol the average Bond enthalpy of N-H bond in Nh3 is ?? Ans is +352 kJ per mole.plz give step.?.
The standard enthalpy of formation of Nh3 is 46.0 kJ per mol if the enthalpy of formation of H2 from its atom is -436 kJ per mol and that of N2 is -712 kJ per mol the average Bond enthalpy of N-H bond in Nh3 is ?? Ans is +352 kJ per mole.plz give step.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The standard enthalpy of formation of Nh3 is 46.0 kJ per mol if the enthalpy of formation of H2 from its atom is -436 kJ per mol and that of N2 is -712 kJ per mol the average Bond enthalpy of N-H bond in Nh3 is ?? Ans is +352 kJ per mole.plz give step.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The standard enthalpy of formation of Nh3 is 46.0 kJ per mol if the enthalpy of formation of H2 from its atom is -436 kJ per mol and that of N2 is -712 kJ per mol the average Bond enthalpy of N-H bond in Nh3 is ?? Ans is +352 kJ per mole.plz give step.?.
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The standard enthalpy of formation of Nh3 is 46.0 kJ per mol if the enthalpy of formation of H2 from its atom is -436 kJ per mol and that of N2 is -712 kJ per mol the average Bond enthalpy of N-H bond in Nh3 is ?? Ans is +352 kJ per mole.plz give step.?, a detailed solution for The standard enthalpy of formation of Nh3 is 46.0 kJ per mol if the enthalpy of formation of H2 from its atom is -436 kJ per mol and that of N2 is -712 kJ per mol the average Bond enthalpy of N-H bond in Nh3 is ?? Ans is +352 kJ per mole.plz give step.? has been provided alongside types of The standard enthalpy of formation of Nh3 is 46.0 kJ per mol if the enthalpy of formation of H2 from its atom is -436 kJ per mol and that of N2 is -712 kJ per mol the average Bond enthalpy of N-H bond in Nh3 is ?? Ans is +352 kJ per mole.plz give step.? theory, EduRev gives you an
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