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Question

Methane is a commercial source of H2 through the reaction
(I) CH4(g)+12O2(g)CO(g)+2H2(g)
Based on the following thermochemical equations (II to IV)
(II) CH4(g)+2O2(g)CO2(g)+2H2O(g); H=800 kJ

(III) CH4(g)+CO2(g)2CO(g)+2H2(g); H=+235 kJ

(IV) CH4(g)+H2O(g)CO(g)+3H2(g); H=+204 kJ
H of equation (I) is

A
-35.75 kJ
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B
- 39.25 kJ
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C
- 25.75 kJ
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D
+ 25.75 kJ
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Solution

The correct option is B - 39.25 kJ
(II + III) gives V
2CH4 (g)+2O2 (g)2CO (g)+2H2 (g)+2H2O (g);
H=800+235=565 kJ
(V) CH4 (g)+O2 (g)CO(g)+H2 (g)+H2O (g);
H=5652kJ=282.5 kJ
Eq. (IV) + Eq. (V) gives
2CH4 (g)+O2 (g)2CO (g)+4H2 (g);
H=(282.5+204) kJ
CH4 (g)+12O2 (g)CO (g)+2H2 (g);
H=78.52=39.25 kJ


Hess’s Law of Constant Heat Summation:
Enthalpy change of a chemical equation is always constant whether the process occurs in a single step or several steps. In other words, the total enthalpy change in a reaction depends only upon the initial reactant and final product not upon the path.

Let say a reaction goes from A to D

AD ΔH

AB ΔH1

BC ΔH2

CD ΔH3

According to the Hess’s Law of Constant Heat Summation

ΔH=ΔH1+ΔH2+ΔH3

  • Chemical equations can be added or subtracted to yield the required equation.

  • Corresponding enthalpies are also manipulated in the same way to get the desired enthalpy.

Example:

C(s)+12O2(g)CO(g) ΔrH

Reaction 1:
C(s)+O2(g)CO2(g) ΔrH1

Reaction 2:
CO(g)+12O2(g)CO2(g) ΔrH2

Now, if we reverse the reaction 2

Reaction 3:
CO2(g)CO(g)+12O2(g) ΔrH2

Now adding equation 1 and equation 3, we get

C(s)+12O2(g)CO(g)

So, ΔrH=ΔrH1ΔrH2


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