Question

The enthalpy of combustion of methane, graphite and dihydrogen at `298 K` are, `-890.3 kJ mol^(-1) -393.5 kJ mol^(-1)`, and `-285.8 kJ mol^(-1)` respectively. Enthapy of formation of `CH_(4)(g)` will be
A. `-74.8 kJ mol^(-1)`
B. `-52.27 kJ mol^(-1)`
C. `+74.8 kJ mol^(-1)`
D. `+52.26 kJ mol^(-1)`

Answer 1

Correct Answer - a
Given:
a. `CH_(4)(g)+2O_(2)(g) rarr CO_(2)(g)+2H_(2)O(l), DeltaH=-890.3 kJ mol^(-1)`
b. `C(s)+O_(2)(g) rarr CO_(2)(g), DeltaH=-393.5 kJ mol^(-1)`
c. `H_(2)(g)+1/2 O_(2)(g)rarr H_(2)O(l), DeltaH=-285.8 kJ mol^(-1)`
Aim: `C(s)+2H_(2)(g) rarr CH_(4)(g), DeltaH=?`
Equation `(b)+2xx` Equation (c ) -Equation (a) gives the required equation with `DeltaH=-393.5+2(-285.8)-(-890.3) kJ mol^(-1)`
`=-74.8 kJ mol^(-1)`