Question

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?

Short answer

The standard enthalpy of formation for nitromethane is: $$ \Delta H_{\mathrm{f\ of\ Nitromethane}}^{\circ} = 1288.5 - (-393.5 +0+ 2 \times (-241.8))\,\text{kJ} = 25.6\, \text{kJ} $$ The mass of nitrogen produced in the reaction is given by: $$ \text{Mass of}\, \mathrm{N}_{2} = n_{\mathrm{N}_{2}}\times (28.02\, \text{g/mol}) = \frac{(P_{\mathrm{N}_{2}}\,\text{torr})(15.0\,\text{L})}{(62.364\,\text{L\ mmHg/mol\ K})(373\,\text{K})} \times (28.02\, \text{g/mol}) $$ Substituting the values, we get: $$ \text{Mass of}\, \mathrm{N}_{2} = \frac{(950\,\text{torr})(0.134)(15.0\,\text{L})}{(62.364\,\text{L\ mmHg/mol\ K})(373\,\text{K})} \times (28.02\, \text{g/mol}) = 10.7\, \text{g} $$

Step-by-step solution

1. Balance the chemical equation

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: $$ \mathrm{CH}_{3} \mathrm{NO}_{2}(l)+\frac{3}{2}\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{N}_{2}(g)+2\mathrm{H}_{2} \mathrm{O}(g) $$

a. Calculate standard enthalpy of formation for nitromethane

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: $$ \Delta H_{\mathrm{f\ of\ Nitromethane}}^{\circ} = -(-1288.5) + \sum \Delta H_{\mathrm{f\ of\ products}}^{\circ} = 1288.5 - (-393.5 +0+ 2 \times (-241.8))\,\text{kJ} $$

2. Calculate mass of nitrogen produced in the reaction

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: $$ \text{Mass of}\, \mathrm{N}_{2} = n_{\mathrm{N}_{2}}\times M_{\mathrm{N}_{2}} = n_{\mathrm{N}_{2}}\times (28.02\, \text{g/mol}) $$

Key concepts

Enthalpy of Formation

In chemistry, the enthalpy of formation refers to the heat change when one mole of a compound forms from its elements in their most stable forms under standard conditions (1 atm and 298 K). This value is denoted by \( \Delta H_{\mathrm{f}}^{\circ} \) and provides insight into the stability and energy changes associated with chemical reactions. For nitromethane, finding the \( \Delta H_{\mathrm{f}}^{\circ} \) involves understanding its role in the overall reaction.
The equation for enthalpy change of a reaction is:

• \( \Delta H_{\mathrm{rxn}}^{\circ} = \sum \Delta H_{\mathrm{f\ of\ products}}^{\circ} - \sum \Delta H_{\mathrm{f\ of\ reactants}}^{\circ} \)

By rearranging this formula, we can calculate \( \Delta H_{\mathrm{f}}^{\circ} \) for any substance if the reaction's enthalpy change is known.
In the reaction involving nitromethane, the enthalpy of the reaction (\( \Delta H_{\mathrm{rxn}}^{\circ} = -1288.5\,\text{kJ} \)) helps us deduce \( \Delta H_{\mathrm{f}}^{\circ} \) for nitromethane by considering the formation enthalpy of the products.
To find this value accurately, we consider the known \( \Delta H_{\mathrm{f}}^{\circ} \) for common products like \( \text{CO}_{2}\) and \( \text{H}_{2}\text{O} \), and apply the formula to solve for nitromethane's enthalpy of formation.

Balanced Chemical Equation

A balanced chemical equation is essential for accurately describing chemical reactions because it ensures that the same number of each type of atom appears on both sides of the equation. This reflects the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.
To balance an equation, adjust the coefficients before each compound until there are equal numbers of each type of atom on both sides.
For example, in the reaction of nitromethane combusting with oxygen:

• \( \mathrm{CH}_{3} \mathrm{NO}_{2}(l)+\frac{3}{2}\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{N}_{2}(g)+2\mathrm{H}_{2} \mathrm{O}(g) \)

In this balanced equation, there is:

• 1 carbon atom, 4 hydrogen atoms, 2 nitrogen atoms, and 5 oxygen atoms on both sides,
• achieving the balance needed for stoichiometry.

Balanced chemical equations are crucial for performing quantitative calculations in reactions, such as determining the enthalpy change or the yield of products.
Using balanced equations helps predict how much reactant is needed or how much product will form, an important consideration when calculating energy changes or substance amounts.

Ideal Gas Law

The Ideal Gas Law is a crucial equation in chemistry for relating the pressure, volume, temperature, and number of moles of a gas. The formula is expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin. This equation helps us understand how gases behave under different conditions.
In the context of the nitromethane reaction, after the combustion process, we deal with gases in a flask, specifically measuring the mass of nitrogen produced.
Here’s how the Ideal Gas Law can be applied:

• Calculate the partial pressure of nitrogen using its mole fraction and total pressure.
• Convert \( P \) and other units to match the gas constant \( R \) in consistent units.
• Use the Ideal Gas Law to find the amount of nitrogen in moles.

Finally, converting moles to mass involves multiplying by the molar mass of nitrogen (28.02 g/mol).
The Ideal Gas Law is instrumental in predicting gas behaviors and allows calculations for the quantities involved, like the mass in this example, getting us from known conditions to desired results.