Chemistry, asked by compengmedia, 5 months ago

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 nidhikumari69032
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 studay07
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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