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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?

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    $\begingroup$ 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. $\endgroup$ – andselisk Jan 24 at 21:53
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    $\begingroup$ 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):) $\endgroup$ – andselisk Jan 24 at 22:08
  • $\begingroup$ Are you at least given the heart of combustion? $\endgroup$ – Chester Miller Jan 24 at 23:14
  • $\begingroup$ @user9988, be sure to update the question with your approach to a solution, to make this question less homework-y. $\endgroup$ – tschoppi Jan 25 at 9:58
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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:

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

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

$$\ce{N2 + 3 H2 -> 2 NH3}$$

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).

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