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Chemistry Question on Enthalpy change

Given, N2(g)+3H2(g)→2NH3(g)N_2(g) + 3H_2(g)→2NH_3(g); ∆rHΘ=−92.4 kJmol–1∆_rH^Θ=-92.4\ kJ mol^{–1}
What is the standard enthalpy of formation of NH3NH_3 gas?

Answer

Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state.
Re-writing the given equation for 1 mole of NH3(g)NH_3(g),
12N2(g)+32H2(g)→NH3(g)\frac 12 N_2(g) + \frac 32H_2(g) → NH_3(g)
Standard enthalpy of formation of NH3NH_3
= 12∆rHΘ\frac 12 ∆_rH^Θ
= 12(–92.4 kJmol–1)\frac 12(–92.4\ kJ mol^{–1})
= –46.2 kJmol–1–46.2\ kJ mol^{–1}