When compression is carried out in infinite steps with change in pressure, it is a reversible process. Work done on the gas is represented by the shaded area.

The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states (reference states) is called standard molar enthalpy of formation, \({ \Delta }_{ f }{ H }^{ o }\) .
\(Ca(s)+C(s)+\frac { 3 }{ 2 } { O }_{ 2 }(g)\longrightarrow Ca{ CO }_{ 3 }(s);{ \Delta }_{ f }{ H }^{ o }\)
This reaction is different from the given reaction.
Hence, \({ \Delta }_{ r }{ H }^{ o }\neq { \Delta }_{ f }{ H }^{ o }\)

Given, \(\frac { 1 }{ 2 } { N }_{ 2 }(g)+{ \frac { 3 }{ 2 } }{ H }_{ 2 }(g)\longrightarrow { NH }_{ 3 }(g);\)
\({ \Delta }_{ f }{ H }^{ \Theta }=-91.8 \ kJ \ { mol }^{ -1 }\)
(\({ \Delta }_{ f }{ H }^{ \Theta }\) means enthalpy of formation of 1 mole of NH₃)
\(\therefore\) Enthalpy change for the formation of 2 moles of NH₃
\({ N }_{ 2 }(g)+{ 3H }_{ 2 }(g)\rightarrow { 2NH }_{ 3 }(g);\)
\({ \Delta }_{ f }{ H }^{ \Theta }=2\times -91.8=-183.6 \ kJ \ { mol }^{ -1 }\)
And for the reverse reaction. \(2{ NH }_{ 3 }(g)\longrightarrow { N }_{ 2 }(g)+{ 3H }_{ 2 }(g);{ \Delta }_{ f }{ H }^{ \Theta }=+183.6 \ kJ \ { mol }^{ -1 }\)  Hence, the value of \({ \Delta }_{ f }{ H }^{ \Theta }\) for NH₃ is +183.6 kJ mol^-1

At constant pressure \(q_{ p }=\triangle U+p\triangle V.\)
But \(\triangle U+p\triangle V=\triangle H\).
\(\therefore \ q_{ p }=\triangle H\). As \(\triangle H\) is a state function, therefore, \(q_{ p }\) is a state function.

At Constant volume By first law of thermodynamics, \(\triangle U=q+W \ or \ q=\triangle U-W.\)But  \(W=-p\triangle V.\) Hence, \(q=\triangle U+p\triangle V.\) But as volume remains constant, \(\triangle V=0\) .
\(\therefore qV=\triangle U\). But \(\triangle U\) is state function. Hence, \(q_{ v }\) is state function.