Nitromethane, \(\mathrm{CH}_{3} \mathrm{NO}_{2},\) can be used as a fuel. When the liquid is burned, the (unbalanced) reaction is mainly $$ \mathrm{CH}_{3} \mathrm{NO}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ a. The standard enthalpy change of reaction $\left(\Delta H_{\mathrm{rxn}}^{\circ}\right)$ for the balanced reaction (with lowest whole-number coefficients \()\) is $-1288.5 \mathrm{kJ} .\( Calculate \)\Delta H_{\mathrm{f}}^{\circ}$ for nitromethane. b. A 15.0 -L flask containing a sample of nitromethane is filled with \(\mathrm{O}_{2}\) and the flask is heated to \(100 .^{\circ} \mathrm{C}\) . 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 \(\left(\chi_{\text { nitrogen }}\right)\) is 0.134 after the reaction is complete, what mass of nitrogen was produced?

First, we need to balance the chemical equation, so that we can find the coefficients of the reactants and products: $$ \mathrm{CH}_{3} \mathrm{NO}_{2}(l)+\frac{3}{2}\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{N}_{2}(g)+2\mathrm{H}_{2} \mathrm{O}(g) $$ So the balanced equation is: $$ \mathrm{CH}_{3} \mathrm{NO}_{2}(l)+\frac{3}{2}\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{N}_{2}(g)+2\mathrm{H}_{2} \mathrm{O}(g) $$

We know that for any reaction, $$ \Delta H_{\mathrm{rxn}}^{\circ} = \sum \Delta H_{\mathrm{f\ of\ products}}^{\circ} - \sum \Delta H_{\mathrm{f\ of\ reactants}}^{\circ} $$ Rearranging the equation to find \(\Delta H_{\mathrm{f\ of\ Nitromethane}}^{\circ}\), $$ \Delta H_{\mathrm{f\ of\ Nitromethane}}^{\circ} = -\Delta H_{\mathrm{rxn}}^{\circ} + \sum \Delta H_{\mathrm{f\ of\ products}}^{\circ} - \Delta H_{\mathrm{f\ of\ O}_{2}}^{\circ} $$ Since the standard enthalpy of formation for oxygen (\(\mathrm{O}_{2}\)) is zero, we get: $$ \Delta H_{\mathrm{f\ of\ Nitromethane}}^{\circ} = -\Delta H_{\mathrm{rxn}}^{\circ} + \sum \Delta H_{\mathrm{f\ of\ products}}^{\circ} $$ We can find the sum of the standard enthalpy of formation for the products \(\left(\sum \Delta H_{\mathrm{f\ of\ products}}^{\circ}\right)\) by looking up their values in a reference or using a standard enthalpy table: $$ \sum \Delta H_{\mathrm{f\ of\ products}}^{\circ} = \Delta H_{\mathrm{f\ of\ CO}_{2}}^{\circ} + \Delta H_{\mathrm{f\ of\ N}_{2}}^{\circ} + 2 \Delta H_{\mathrm{f\ of\ H}_{2}\mathrm{O}}^{\circ} = (-393.5 + 0 + 2 \times (-241.8))\,\text{kJ} $$ Now, we can plug in the values to find the standard enthalpy of formation for nitromethane: $$ \Delta H_{\mathrm{f\ of\ Nitromethane}}^{\circ} = -(-1288.5) + \sum \Delta H_{\mathrm{f\ of\ products}}^{\circ} = 1288.5 - (-393.5 +0+ 2 \times (-241.8))\,\text{kJ} $$

First, let's determine the partial pressure of nitrogen in the flask: Partial pressure of nitrogen = Total pressure × Mole fraction of nitrogen $$ P_{\mathrm{N}_{2}} = (950\,\text{torr})(0.134) $$ Next, we need to convert the partial pressure of nitrogen into moles using the ideal gas law equation, \(PV = nRT\). Make sure to convert all units to the correct SI units: $$ n_{\mathrm{N}_{2}} = \frac{P_{\mathrm{N}_{2}}V}{RT} = \frac{(P_{\mathrm{N}_{2}}\,\text{torr})(15.0\,\text{L})}{(62.364\,\text{L\ mmHg/mol\ K})(373\,\text{K})} $$ Then, we can find the mass of nitrogen produced, using its molar mass: $$ \text{Mass of}\, \mathrm{N}_{2} = n_{\mathrm{N}_{2}}\times M_{\mathrm{N}_{2}} = n_{\mathrm{N}_{2}}\times (28.02\, \text{g/mol}) $$