Question

The source of oxygen that drives the internal combustion engine in an automobile is air. Air is a mixture of gases, which are principally \(\mathrm{N}_{2}(\sim 79 \%)\) and \(\mathrm{O}_{2}(\sim 20 \%)\). In the cylinder of an automobile engine, nitrogen can react with oxygen to produce nitric oxide gas, NO. As NO is emitted from the tailpipe of the car, it can react with more oxygen to produce nitrogen dioxide gas. (a) Write balanced chemical equations for both reactions. (b) Both nitric oxide and nitrogen dioxide are pollutants that can lead to acid rain and global warming; collectively, they are called "NOx" gases. In 2004, the United States emitted an estimated 19 million tons of nitrogen dioxide into the atmosphere. How many grams of nitrogen dioxide is this? (c) The production of \(\mathrm{NO}_{\mathrm{x}}\) gases is an unwanted side reaction of the main engine combustion process that turns octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\) into \(\mathrm{CO}_{2}\) and water. If \(85 \%\) of the oxygen in an engine is used to combust octane, and the remainder used to produce nitrogen dioxide, calculate how many grams of nitrogen dioxide would be produced during the combustion of 500 grams of octane.

Short answer

The balanced chemical equations for the reactions are: 1) N2 + O2 -> 2NO 2) 2NO + O2 -> 2NO2 There were about \(1.9 \times 10^{13}\) grams of nitrogen dioxide emitted in the United States in 2004. During the combustion of 500 grams of octane, approximately 754.4 grams of nitrogen dioxide would be produced.

Step-by-step solution

Balancing Chemical Equations

We need to write balanced chemical equations for the two reactions mentioned in the problem: 1) The reaction of nitrogen with oxygen to produce nitric oxide gas (NO): N2 + O2 -> 2NO 2) The reaction of nitric oxide with oxygen to produce nitrogen dioxide gas (NO2): 2NO + O2 -> 2NO2

Converting Tons to Grams

To find the mass of nitrogen dioxide in grams metric tons needs to be converted to grams. Use the conversion: 1 ton = 1,000,000 grams. 19 million tons = 19,000,000 tons 19,000,000 tons * 1,000,000 grams / ton = 1.9 * 10^13 grams (of nitrogen dioxide)

Combustion of Octane and Nitrogen Dioxide Production

First, we need to write a balanced equation for the combustion of octane: C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O Now, we know that 85% of the oxygen in an engine is used to combust octane, and the remainder (15%) is used to produce nitrogen dioxide. We will first find the moles of octane and then find the moles of oxygen required. Molar mass of octane (C8H18) = (12.01 * 8) + (1.01 * 18) = 96.08 + 18.18 = 114.26 g/mol We are given 500 grams of octane, so the number of moles of octane: moles of octane = 500 g / 114.26 g/mol = 4.37 mol From the balanced equation of octane combustion, we see that 1 mol of octane requires 12.5 mol of O2. Total moles of O2 required for complete combustion of octane: 4.37 mol * 12.5 = 54.63 mol Since 85% of the oxygen is used for octane combustion, we can now find out how much oxygen was used for the NO2 production: 15% of total oxygen needed: 0.15 * 54.63 mol = 8.195 mol Now to find moles of NO2 produced, we can look at the balanced equation for NO2 production (2NO + O2 -> 2NO2). We see that 1 mol of O2 produces 2 mol of NO2. Moles of NO2 produced: 8.195 mol * 2 = 16.39 mol To find the mass of NO2 produced, multiply the moles by the molar mass of NO2: Molar mass of NO2 = (14.01) + (16.00 * 2) = 46.01 g/mol Mass of NO2 produced = 16.39 mol * 46.01 g/mol = 754.4 g (of nitrogen dioxide)

Key concepts

Chemical Equations Balancing

When chemistry students first encounter the combustion of octane, they quickly learn the significance of balancing chemical equations. A balanced chemical equation represents the conservation of mass; the number of atoms of each element on the reactant side must equal the number on the product side. For the combustion of octane (\textbf{C8H18}), it is imperative to precisely balance this equation to understand how much oxygen (\textbf{O2}) is required for the process.

To illustrate, the balanced equation looks like this:\[ \text{C8H18} + 12.5 \text{O2} \rightarrow 8\text{CO2} + 9 \text{H2O} \]
In this equation, we can observe that each carbon (C) and hydrogen (H) atom from the octane is accounted for in the products: carbon dioxide (\textbf{CO2}) and water (\textbf{H2O}). Notice the coefficient 12.5 in front of \textbf{O2}, which tells us that 12.5 moles of oxygen are necessary for every mole of octane. Balancing chemical equations is not only fundamental to comprehend the stoichiometry of the reaction but is also vital to ensure that calculations affecting environmental outcomes, such as production of NOx gases, are accurate.

NOx Gases and Pollution

Nitric oxide (\textbf{NO}) and nitrogen dioxide (\textbf{NO2}) are known as 'NOx gases' and represent a serious environmental hazard. These pollutants are byproducts of the high-temperature combustion that occurs in car engines. Their primary modes of formation are the reactions between nitrogen (\textbf{N2}) and oxygen (\textbf{O2}) present in the air.\[ \text{N2} + \text{O2} \rightarrow 2\text{NO} \]followed by\[ 2\text{NO} + \text{O2} \rightarrow 2\text{NO2} \]
NOx gases contribute to several environmental issues, including acid rain, smog, and have implications for human health as they can lead to respiratory problems. Beyond this, they play a role in forming ground-level ozone and in the greenhouse effect, potentially contributing to climate change. Taking the 2004 data from the United States, an astonishing 19 million tons of nitrogen dioxide were emitted. To visualize this staggering amount, it's translated to grams:\[ 19,000,000 \text{ tons} \times 1,000,000 \text{ grams/ton} = 1.9 \times 10^{13} \text{ grams} \]
Understanding the vast scale of these emissions underscores the importance of reducing NOx gas production and the impact vehicles have on our environment.

Stoichiometry Calculations

Stoichiometry is the part of chemistry that involves calculating the amounts of reactants and products in a chemical reaction. For instance, if we want to calculate the amount of nitrogen dioxide produced during the combustion of octane, we must rely on stoichiometry. The process starts with understanding the balanced equation and identifying the molar ratios between reactants and products. This leads us to calculate the moles of all substances involved based on the given masses.Take the example of burning 500 grams of octane, where only 15% of the oxygen is responsible for NO2 production. The mole calculation would look like this:\[ \text{Moles of octane} = \frac{500 \text{ g}}{114.26 \text{ g/mol}} = 4.37 \text{ mol} \]This allows us to use the molecular ratios from the balanced equations to compute the moles of oxygen used and consequently the moles of NO2 produced:\[ \text{Mass of NO2 produced} = 16.39 \text{ mol} \times 46.01 \text{ g/mol} = 754.4 \text{ g} \]
Through stoichiometry, we can predict quantities of unwanted pollutants and strategize on how to minimize their release, playing a pivotal role in both the chemistry and the environmental science fields.