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ΔfU° of formation of CH4(g) at a certain temperature is - 393 kJ mol-1. The value of ΔfH° is
- a)zero
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
ΔfU° of formation of CH4(g) at a certain temperature is - 39...
The balanced equation for combustion of methane is:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)Here, Δng=1−3=−2
ΔH0=ΔU0+ΔngRT
ΔH0 =−393−2RT
∴ΔH0<ΔU0
CH4(g)+2O2(g)→CO2(g)+2H2O(l)Here, Δng=1−3=−2
ΔH0=ΔU0+ΔngRT
ΔH0 =−393−2RT
∴ΔH0<ΔU0
Most Upvoted Answer
ΔfU° of formation of CH4(g) at a certain temperature is - 39...
The balanced equation for combustion of methane is:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)Here, Δng=1−3=−2
ΔH0=ΔU0+ΔngRT
ΔH0 =−393−2RT
∴ΔH0<ΔU0
CH4(g)+2O2(g)→CO2(g)+2H2O(l)Here, Δng=1−3=−2
ΔH0=ΔU0+ΔngRT
ΔH0 =−393−2RT
∴ΔH0<ΔU0
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Community Answer
ΔfU° of formation of CH4(g) at a certain temperature is - 39...
Standard enthalpy of formation is what we get
when we do all of the steps needed to
dissociate elements into individual atoms, then
ionizing them so they will form the lattice
material at a later stage. Lattice energy is just
one of the steps involved in the whole process,
and is the step we do when joining the ionized
elements together in a crystal. The ionized
forms of the elements are at a different energy
relative to the original forms they started in, so
that is why lattice energy is different compared
to enthalpy of formation. Processes like ionization and dissociation require energy to be supplied( del D, etc. = +ve) while lattice formation involves release of energy(del U = -ve). Thus, in general, the magnitude of del H (= del U + del D +.....) is less than del U but it's value is higher than del U.
Probably this can be the explanation. Hope that helps.!
when we do all of the steps needed to
dissociate elements into individual atoms, then
ionizing them so they will form the lattice
material at a later stage. Lattice energy is just
one of the steps involved in the whole process,
and is the step we do when joining the ionized
elements together in a crystal. The ionized
forms of the elements are at a different energy
relative to the original forms they started in, so
that is why lattice energy is different compared
to enthalpy of formation. Processes like ionization and dissociation require energy to be supplied( del D, etc. = +ve) while lattice formation involves release of energy(del U = -ve). Thus, in general, the magnitude of del H (= del U + del D +.....) is less than del U but it's value is higher than del U.
Probably this can be the explanation. Hope that helps.!
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ΔfU° of formation of CH4(g) at a certain temperature is - 393 kJ mol-1. The valueof ΔfH° isa)zerob)c)d)Correct answer is option 'C'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about ΔfU° of formation of CH4(g) at a certain temperature is - 393 kJ mol-1. The valueof ΔfH° isa)zerob)c)d)Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ΔfU° of formation of CH4(g) at a certain temperature is - 393 kJ mol-1. The valueof ΔfH° isa)zerob)c)d)Correct answer is option 'C'. Can you explain this answer?.
ΔfU° of formation of CH4(g) at a certain temperature is - 393 kJ mol-1. The valueof ΔfH° isa)zerob)c)d)Correct answer is option 'C'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about ΔfU° of formation of CH4(g) at a certain temperature is - 393 kJ mol-1. The valueof ΔfH° isa)zerob)c)d)Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ΔfU° of formation of CH4(g) at a certain temperature is - 393 kJ mol-1. The valueof ΔfH° isa)zerob)c)d)Correct answer is option 'C'. Can you explain this answer?.
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