Using Hess's law to find molar enthalpy of formation without an initial reaction I'm having issues relating Hess's law to this question:


  The given chemical equation represent the combustion of ammonia and the combustion of hydrogen
  
  $$
\begin{align}
\ce{4 NH3 + 3 O2 &amp;-&gt;6 H2O + 2N2} &amp;\qquad &amp;ΔH_1 = \pu{-1516 kJ} \tag{1} \\
\ce{2 H2 + O2 &amp;-&gt; 2 H2O} &amp;\qquad &amp;ΔH_2 = \pu{-572 kJ} \tag{2} \\
\end{align}
$$
  
  What is the molar enthalpy of formation for ammonia?
  
  A. $\pu{-100 kJ}$; B. $\pu{-50 kJ}$; C. $\pu{50 kJ}$; D. $\pu{100 kJ}$.


I'm not sure how I would apply the  product–reactant equation here, or rearrange the equation to get my answer since I don't have an initial reaction to compare it to. How would I go about answering this question?
Since you are asked to find the enthalpy of formation for ammonia, it's convenient to write the equation of ammonia synthesis normalized for $\pu{1 mol}$ of $\ce{NH3}$:

$$\ce{0.5 N2 + 1.5 H2 -&gt; NH3}$$

To obtain this equation and apply Hess's law, the following linear combination of the provided equations should be used:

$$
\begin{align}
\ce{4 NH3 + 3 O2 &amp;-&gt;6 H2O + 2N2} &amp; ΔH_1 &amp;= \pu{-1516 kJ} &amp;&amp;|\cdot(-0.25) \tag{1} \\
\ce{2 H2 + O2 &amp;-&gt; 2 H2O} &amp; ΔH_2 &amp;= \pu{-572 kJ} &amp;&amp;|\cdot 0.75 \tag{2} \\
\hline
\ce{0.5 N2 + 1.5 H2 &amp;-&gt; NH3} &amp; ΔH_3 &amp;= -0.25ΔH_1 + 0.75ΔH_2 \tag{3}
\end{align}
$$

So the final answer is

$$ΔH_3 = -0.25\cdot(\pu{-1516 kJ}) + 0.75\cdot(\pu{-572 kJ}) = \pu{-50 kJ}$$

The catch here is that you are asked to find molar enthalpy of formation, and calculating for the "convenient" equation with integer coefficients

$$\ce{N2 + 3 H2 -&gt; 2 NH3}$$

you eventually would obtain the doubled value of $\pu{-100 kJ mol-1}$ referred to $\pu{2 mol}$ of ammonia (I almost fell for it, too).