# How do I calculate the standard enthalpy of formation (ΔH∘f) for nitromethane?

Jul 7, 2017

I got $\textcolor{b l u e}{- \text{47 kJ/mol}}$, assuming a constant-pressure calorimeter. So, I answer the question as-written...

The "correct" answer marked by MasteringChemistry was $- \text{401.6 kJ/mol}$, but that is based on a $\Delta {H}_{C}^{\circ}$ that doesn't account for how the reaction given here uses $2$ mols of nitromethane, rather than $1$.

(Note: The NIST reference I cited below uses a constant-volume calorimeter with a small difference between $\Delta {E}_{C}^{\circ}$ and $\Delta {H}_{C}^{\circ}$, so it is a different situation!)

I think your question isn't totally clear. Some other websites claim the water formed is a liquid, but in order to get $\Delta {H}_{C}^{\circ}$, you need a constant-pressure calorimeter, which is open to the atmosphere.

That indicates the water formed should be a gas, as the container is not rigid and closed, and condensation is not induced.

(If it were, then the question should have asked for the standard internal energy of combustion, $\Delta {E}_{C}^{\circ}$, to be more accurate.)

2CH_3NO_2(l)+3/2O_2(g)→2CO_2(g)+3H_2Ocolor(red)((g))+N_2(g)

And you were given

$\Delta {H}_{C}^{\circ} = - \text{709.2 kJ/mol}$,

which is the same as the enthalpy of reaction for a combustion reaction, forming ONE mol of nitromethane. This is given in general by:

CH_3NO_2(l)+3/4O_2(g)→CO_2(g)+3/2H_2O(g)+1/2N_2(g)

$\Delta {H}_{r x n}^{\circ} = {\sum}_{P} {\nu}_{P} \Delta {H}_{f , P}^{\circ} - {\sum}_{R} {\nu}_{R} \Delta {H}_{f , R}^{\circ}$

$= - \text{709.2 kJ/mol}$,

where $P$ and $R$ stand for products and reactants, respectively, and $\nu$ is the stoichiometric coefficient.

And here, you are asked to find $\Delta {H}_{f , C {H}_{3} N {O}_{2} \left(l\right)}^{\circ}$. From your appendix, you can look these up:

$\Delta {H}_{f , {O}_{2} \left(g\right)}^{\circ} = \text{0 kJ/mol}$

$\Delta {H}_{f , C {O}_{2} \left(g\right)}^{\circ} = - \text{393.5 kJ/mol}$

$\Delta {H}_{f , {H}_{2} O \left(g\right)}^{\circ} = - \text{241.8 kJ/mol}$

$\Delta {H}_{f , {N}_{2} \left(g\right)}^{\circ} = \text{0 kJ/mol}$

When you realize that the reaction utilizes TWO mols of nitromethane, that's going to actually give you, for this reaction as-written:

2(-"709.2 kJ/mol") = overbrace([2DeltaH_(f,CO_2(g))^@ + 3DeltaH_(f,H_2O(g))^@ + 1(0)])^"Products" - overbrace([2DeltaH_(f,CH_3NO_2(l))^@ + 3/2 (0)])^"Reactants"

Now just solve for $\Delta {H}_{f , C {H}_{3} N {O}_{2} \left(l\right)}^{\circ}$ as the only unknown.

$\textcolor{b l u e}{\Delta {H}_{f , C {H}_{3} N {O}_{2} \left(l\right)}^{\circ}} = \frac{2 \times \left(- \text{709.2 kJ/mol}\right) - \left[2 \Delta {H}_{f , C {O}_{2} \left(g\right)}^{\circ} + 3 \Delta {H}_{f , {H}_{2} O \left(g\right)}^{\circ}\right]}{- 2}$

= color(blue)(???)

I get $\textcolor{b l u e}{- \text{47 kJ/mol}}$. The factor of $2$ in front of the $- 709.2$ makes the difference between $- 47$ and $- 401.6$ $\text{kJ/mol}$.

(If you had used the $\Delta {H}_{f}^{\circ}$ for liquid water, you would get $- \text{113 kJ/mol}$, which is in the NIST reference, but the reference uses a bomb calorimeter, which is a constant-volume system that enforces formation of liquid water, not water vapor.)