KEAM 2019 Physics and Chemistry Paper

Given : C+2H2⟶CH4
∆fH∘=−76.2kJmol−1......(i)
C+O2⟶CO2
∆fH∘=−349.8kJmol−1.....(ii)
H2+‌

1

2

O2⟶H2O(l)
H2O(l)⟶H2O(g)
∆vH∘=44kJmol−1.....(iv)
Oxidation of CH4 into its gaseous products is given as
CH4+2O2⟶CO2+2H2O∆H=?
To get the required equation, rewrite the above equation.
CH4⟶C+2H2
∆fH∘=+76.2kJmol−1.....(i)
C+O2⟶CO2
∆fH∘=−394.8kJmol−1.....(ii)
2H2+O2⟶2H2O(l)
∆fH∘=2×−285.82kJmol−1‌‌.......(iii)
2H2O(l)⟶H2O(g)
∆vH∘=2×44kJmol−1......(iv)
given ∆H=−80224
Thus, heat evolved when 22.4L (or 1 mole of CH4 ) is oxidised into its gaseous products is 802.64kJ.