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 \(15.0\) - \(\mathrm{L}\) 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?

To find the standard enthalpy of formation, we first need to balance the given equation: \[ \mathrm{CH}_{3} \mathrm{NO}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{N}_{2}(g)+2\mathrm{H}_{2} \mathrm{O}(g) \] Balanced equation: \[ 2\mathrm{CH}_{3} \mathrm{NO}_{2}(l)+3\mathrm{O}_{2}(g) \longrightarrow 2\mathrm{CO}_{2}(g)+2\mathrm{N}_{2}(g)+6\mathrm{H}_{2} \mathrm{O}(g) \]

We can use the standard enthalpy of formation relationship to find the value of the enthalpy change for nitromethane: \[ \Delta H_{rxn}^{\circ}=\sum n \Delta H_{f(products)}^{\circ}-\sum n \Delta H_{f(reactants)}^{\circ} \] where \(\Delta H_{rxn}^{\circ}\) is the standard enthalpy change of reaction, \(\Delta H_{f(products)}^{\circ}\) and \(\Delta H_{f(reactants)}^{\circ}\) are the standard enthalpy of formation for products and reactants, and n represents the coefficients in the balanced equation. We know the standard enthalpy of formation for the products and reactants excluding nitromethane: \(ΔH_{f(CO_2)}^{\circ}=-393.5 kJ \cdot mol^{-1}\) \(ΔH_{f(N_2)}^{\circ}=0 kJ \cdot mol^{-1}\) \(ΔH_{f(H_2O)}^{\circ}=-241.8 kJ \cdot mol^{-1}\) \(ΔH_{f(O_2)}^{\circ}=0 kJ \cdot mol^{-1}\) We can rewrite the relationship to isolate \(\Delta H_{f(\mathrm{CH}_{3}\mathrm{NO}_{2})}^{\circ}\): \[ \Delta H_{f(\mathrm{CH}_{3} \mathrm{NO}_{2})}^{\circ}=\frac{\Delta H_{rxn}^{\circ} -\sum n \Delta H_{f(products)}^{\circ}+\sum n \Delta H_{f(reactants)}^{\circ}}{2} \] Plugging in the known values: \[ \Delta H_{f(\mathrm{CH}_{3} \mathrm{NO}_{2})}^{\circ}=\frac{-1288.5 \,\text{kJ}- [(2\cdot(-393.5\,\text{kJ})+(2(0\,\text{kJ}))+6\cdot(-241.8\,\text{kJ}))-3\cdot(0\,\text{kJ})]}{2} \] Calculating the standard enthalpy of formation for nitromethane: \[ \Delta H_{f(\mathrm{CH}_{3} \mathrm{NO}_{2})}^{\circ}=-93.65\,\text{kJ/mol} \] #Part b: Mass of nitrogen produced#

The mole fraction of nitrogen is given as 0.134. We can use this value to find the partial pressure of nitrogen in the gas mixture: \[ P_{\text{nitrogen}}=\chi_{\text {nitrogen }} \times P_{\text{total}} \] Plugging in the given values: \[ P_{\text{nitrogen}}=0.134 \times 950\, \text{torr} \] Calculating the partial pressure of nitrogen: \[ P_{\text{nitrogen}}=127.3\, \text{torr} \]

We can apply the ideal gas law to determine the moles of nitrogen: \[ PV=nRT \\ n=\frac{PV}{RT} \] where P is pressure in atm, V is volume, n is the moles of the gas, R is the gas constant, and T is the temperature in Kelvin. Convert the given temperature of \(100^{\circ}\) C to Kelvin and the pressure in torr to atm: \(T_{K}=100+273.15=373.15\, \text{K}\) \(P_{atm}=\frac{127.3\, \text{torr}}{760\, \text{torr/atm}}=0.1675\, \text{atm}\) Plugging in the known values into the ideal gas law: \[ n=\frac{(0.1675\,\text{atm})(15.0\,\mathrm{L})}{(0.0821\, \mathrm{L\cdot atm/mol\cdot K})(373.15\, \mathrm{K})} \] Calculate the moles of nitrogen produced: \[ n=0.108\, \text{mol} \] Now, convert the moles of nitrogen into mass: \[ \text{mass}=\text{moles} \times \text{molar mass} \] The molar mass of nitrogen is 28.02 g/mol, therefore: \[ \text{mass}=(0.108\, \text{mol})(28.02\, \text{g/mol}) \] Calculate the mass of nitrogen: \[ \text{mass}=3.02\, \text{g} \] The mass of nitrogen produced is 3.02 g.