The reaction for the combustion of carbon into CO₂ is
C(s) + O₂(g) \(\longrightarrow \) CO₂(g); \(\triangle H\) = -393.5 KJ mol^-1
Heat released in the formation of 44 g CO₂ = 393.5 KJ
\(\therefore \) Heat released in the formation of 35.2 g CO₂
\(=\frac { 393 \ KJ\times 35.2 \ g }{ 44 \ g } \) = 314.8 KJ

It is a spontaneous process because although \(\Delta H=0\), i.e., energy factor has no role to play but randomness increases, i.e. randomness factor favours the process.

In CH₄ there are four C-H bonds. The enthalpy of atomisation of 1 mole of CH₄ means dissociation of four moles of C-H bond.
\(\therefore \) C-H bond energy per mol = \(\frac { 1665 \ KJ }{ 4 \ mol } \) = 416.25 KJ mol^-1.

\(\Delta _{ r }G=\Delta _{ r }G^{ \circ }+RT \ ln \ K\)
At equilibrium, 
\(0=\Delta _{ r }G^{ \circ }+RT \ ln \ K \ (\because \Delta _{ r }G=0)\)
\( or \ \Delta _{ r }{ G }^{ \circ }=-RT \ ln \ K;\)
\(\Delta _{ r }{ G }^{ \circ }=0 \ when \ K=1\)
For all other values of K, \(\Delta _{ r }{ G }^{ \circ }\) will be non-zero.

\(\Delta _{ r }G^{ \circ }=-2.303 \ RT \ logK_{ p }\)
Here K_p= 0.98, i.e. K_p < 1 therefore, \(\Delta _{ r }G^{ \circ }\) is positive. Hence, the reaction is non-spontaneous.