The standard enthalpy of formation of gaseous H2O at 298 K is –241.82 kJ mol. Estimate its value of 100°C
given the following values of the molar heat capacities at constant pressure:
H2O(g): 35.58 JK- mol H2(9): 28.84 J mol K and 0,(9): 29.37 J mol K-'. Assume heat capacity to
be independent of temperature,
(1) - 272.6 kJ mol-1
(2) 300 kJ mol 1
(3) 242.6 kJ mol-1
(4) 206 kJ mol-1
Answers
Answered by
1
Answer:
Answer is (3)242.6kj mol-1
HormiPolice:
Aw man you had the audacity to not explain the one of the few very clear question
Answered by
4
Answer:
Given = H2O at 298K
ΔH °= -241.82KJ mol
ΔH= H2O(g): 35.58 JK- mol
ΔH= H2(9): 28.84 J mol K
ΔH=O2= 29.37 J mol K-'
to find = standard enthalpy
solution=
we can apply the kirchhoffs law
ΔrH°T₂ - ΔrH°T₁ = ΔrC°p[T₂ - T₁]
but the value of ΔrC°p is not given so,
for the
H2(g) + 1/2O₂(g) →H₂O"(g) = -10.945
now put the given value in above formula,
ΔrH°T₂-(241.82) = [-10.94J (373-298) / 100]
ΔrH°T₂-(241.82) = [-10.94J ( 0.75)]
ΔrH°T₂-(241.82) = 8.205
ΔrH°T₂= -242.6 KJ/mol
so the option 3 is correct answer.
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