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Question

Calculate ΔH at 358 K for the reaction,
Fe2O3(s)+3H2(g) 2Fe(s)+3H2O(l)

Given that , ΔH298=33.29 kJ mol1 and CP for Fe2O3(s), Fe(s), H2O(l) and H2(g) are 103.8, 25.1, 75.3, and 28.8 J/K mol.

A
28.136 kJ/mol
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B
58.136 kJ/mol
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C
28.136 kJ/mol
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D
58.136 kJ/mol
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Solution

The correct option is A 28.136 kJ/mol
Reaction taking place is,
Fe2O3(s)+3H2(g) 2Fe(s)+3H2O(l)
for this,
Cp=(Cp)product(Cp)reactant
putting the values,
ΔCP=[2×25.1+3×75.3][103.8+3×28.8]
=85.9 J/Kmol

We have, from Kirchoff equation,
ΔH2ΔH1T2T1=ΔCP
putting the values,
ΔH358(33290)358298=85.9
ΔH358=28136 J/mol
=28.136 kJ/mol



Theory:

Kirchoff’s Equation:
Kirchhoff's Law describes the enthalpy of a reaction's variation with temperature changes. The overall enthalpy of the reaction will change if the increase in the enthalpy of products and reactants is different.

Kirchoff’s equation can be written as:

If Cp,m is given as a function of temperature,

ΔrH2=ΔrH1+ T2T1ΔrCp,mdT

Where T1 and T2 are in Kelvin (K)




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