Heat is independent of path under two conditions :
By first law of thermodynamics,
$\text{q}=\Delta\text{U}+(-\text{w})$
and, $(-\text{w})=\text{P}\Delta\text{V}$
Therefore, $\text{q}=\Delta\text{U}+\text{P}\Delta\text{V}$
By first law of thermodynamics $\Delta\text{V}=0$
or, $\text{qV}=\Delta\text{U}+0$
or, $\text{qV}=\Delta\text{U}=$ change in internal energy
$\text{q}_\text{p}=\Delta\text{U}+\text{P}\Delta\text{V}$
But, $\Delta\text{U}+\text{P}\Delta\text{V}=\Delta\text{H}$
Therefore, $\text{q}_\text{p}=\Delta\text{H}=$ change in enthalpy
So, at constant volume and at constant pressure heat change is a state function because it is equal to change in internal energy and change in enthalpy respectively which are state functions.
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(Pressure is increased)
$\Delta\text{U}^\ominus$of combustion of methane is – X kJ mol–1. The value of $\Delta\text{H}^\ominus$ is:
$=\Delta\text{U}^\ominus$
$>\Delta\text{U}^\ominus$
$<\Delta\text{U}^\ominus$
$=0$