Question

Nitromethane, CH₃NO₂, can be used as a fuel. When the liquid is burned, the (unbalanced) reaction is mainly

$C{H}_{3}N{O}_2\left(l\right)+O_2\left(g\right)\to C{O}_2\left(g\right)+N_2\left(g\right)+H_{2}O\left(g\right)$

a. The standard enthalpy change of reaction ($\Delta H_{rxn}^o$) for the balanced reaction (with lowest whole-number coefficients) is-1288.5 kJ. Calculate$\Delta H_f^o$ for nitromethane.

b. A 15.0-L flask containing a sample of nitromethane is filled with O2 and the flask is heated to 100.ºC. At this temperature, and after the reaction is complete, the total pressure of all the gases inside the flask is 950. torr. If the mole fraction of nitrogen (Xnitrogen) is 0.134 after the reaction is complete, what mass of nitrogen was produced?

Short answer

1. The for nitromethane is -111.99 kJ/mol.
2. The mass of nitrogen is 2.30 g.

Step-by-step solution

Determination of Hfo  for nitromethane

The balanced equation for the given reaction is,

$2C{H}_{3}N{O}_2\left(l\right)+\frac{3}{2}{O}_2\left(g\right)\to 2C{O}_2\left(g\right)+N_2\left(g\right)+3H_{2}O\left(g\right)$

The $∆H_f^o$for nitromethane is,

$\begin{array}{rcl}\Delta H°& =& \Delta H_f°\left(product\right)-\Delta H_f°\left(reac\tan t\right)\\ \Delta H°& =& \left[2\times \Delta H_f°\left(C{O}_2\right)+3\times \Delta H_f°\left(H_{2}O\right)+\Delta H_f°\left(N_2\right)\right]-\left[2\times \Delta H_f°\left(C{H}_{3}N{O}_2\right)+\frac{3}{2}\times \Delta H_f°\left(O_2\right)\right]\\ -1288.5kJ& =& \left[2\times \left(-393.5kJ\right)+3\times \left(-241.82kJ\right)+\left(0\right)\right]-\left[2\times \Delta H_f°\left(C{H}_{3}N{O}_2\right)+\left(\frac{3}{2}\times 0\right)\right]\\ \Delta H_f°\left(C{H}_{3}N{O}_2\right)& =& -111.99kJ/mol\\ & & \end{array}$

The$∆H$for nitromethane is -111.99 kJ/mol.

Determination of mass of nitrogen

The partial pressure of nitrogen is,

$\begin{array}{rcl}\mathit{P}\mathit{a}\mathit{r}\mathit{t}\mathit{i}\mathit{a}\mathit{l}\mathbf{ }\mathit{p}\mathit{r}\mathit{e}\mathit{s}\mathit{s}\mathit{u}\mathit{r}\mathit{e}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{n}\mathit{i}\mathit{t}\mathit{r}\mathit{o}\mathit{g}\mathit{e}\mathit{n}& \mathbf{=}& \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathbf{ }\mathit{f}\mathit{r}\mathit{a}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{\times }\mathit{t}\mathit{o}\mathit{t}\mathit{a}\mathit{l}\mathbf{ }\mathit{p}\mathit{r}\mathit{e}\mathit{s}\mathit{s}\mathit{u}\mathit{r}\mathit{e}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{134}\mathbf{\times }\mathbf{950}\mathit{t}\mathit{o}\mathit{r}\mathit{r}\\ & \mathbf{=}& \mathbf{127}\mathbf{.}\mathbf{3}\mathit{t}\mathit{o}\mathit{r}\mathit{r}\\ & & \end{array}$

Moles of nitrogen are,

$$\begin{gathered} \mathit{n}\mathbf{=}\frac{\mathbf{P}\mathbf{V}}{\mathbf{R}\mathbf{T}} \\ \mathit{n}\mathbf{=}\frac{\mathbf{127}\mathbf{.}\mathbf{3}\mathbf{t}\mathbf{o}\mathbf{r}\mathbf{r}\mathbf{\times }\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{L}}{\left(62.36367Ltorr{K}^{-1}mo{l}^{-1}\right)\mathbf{\times }\mathbf{373}\mathbf{K}} \\ \mathit{n}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{082}\mathit{m}\mathit{o}\mathit{l} \end{gathered}$$

Now, the mass of nitrogen is,

$\begin{array}{rcl}\mathit{M}\mathit{a}\mathit{s}\mathit{s}& \mathbf{=}& \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{\times }\mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathbf{ }\mathit{m}\mathit{a}\mathit{s}\mathit{s}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{082}\mathit{m}\mathit{o}\mathit{l}\mathbf{\times }\mathbf{28}\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\\ & \mathbf{=}& \mathbf{2}\mathbf{.}\mathbf{30}\mathit{g}\\ & & \end{array}$

Thus, the mass of nitrogen is 2.30 g.