# Using Hess's law to find molar enthalpy of formation without an initial reaction [closed]

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 &->6 H2O + 2N2} &\qquad &ΔH_1 = \pu{-1516 kJ} \tag{1} \\ \ce{2 H2 + O2 &-> 2 H2O} &\qquad &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?

• There is all the info you need. I suggest you start with writing an actual equation for the synthesis of ammonia and then use linear combination of the provided equations. – andselisk Jan 24 at 21:53
• Hold on a sec, that's for 2 mols of $\ce{NH3}$, I think you forgot to divide it in half (it should be around -50 kJ/mol):) – andselisk Jan 24 at 22:08
• Are you at least given the heart of combustion? – Chet Miller Jan 24 at 23:14
• @user9988, be sure to update the question with your approach to a solution, to make this question less homework-y. – tschoppi Jan 25 at 9:58

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 $$\ce{NH3}$$:

$$\ce{0.5 N2 + 1.5 H2 -> NH3}$$

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

$$ΔH_3 = -0.25\cdot(\pu{-1516 kJ}) + 0.75\cdot(\pu{-572 kJ}) = \pu{-50 kJ}$$
$$\ce{N2 + 3 H2 -> 2 NH3}$$
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).