The correct option is B -46.1 kJ/mol
Given, N2(g)+3H2(g)→2NH3(g) ΔHr=−92.2 kJ
Here 2 moles of ammonia are forming but according to definition of enthalpy of formation we know it is for formation of 1 mole of substance. i.e for formation of 1 mole of ammonia
dividing whole equation by 2 we get
12N2(g)+32H2(g)→NH3(g)
ΔHf(NH3)=−92.22=−46.1 kJ
hence, heat of formation of ammonia is −46.1 kJ/mol