1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
- 39.25 kJ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
- 25.75 kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
+ 25.75 kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
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 A→D ΔH A→B ΔH1 B→C ΔH2 C→D 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=ΔrH1−ΔrH2

Suggest Corrections
2
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program