Question

A steel cylinder contains 5.00 mole of graphite (pure carbon) and 5.00 moles of \(\mathrm{O}_{2} .\) The mixture is ignited and all the graphite reacts. Combustion produces a mixture of \(\mathrm{CO}\) gas and \(\mathrm{CO}_{2}\) gas. After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by 17.0\(\%\) . Calculate the mole fractions of \(\mathrm{CO}, \mathrm{CO}_{2},\) and \(\mathrm{O}_{2}\) in the final gaseous mixture.

Short answer

The mole fractions of CO, CO\(_2\), and O\(_2\) after the combustion are approximately 0.291, 0.564, and 0.145, respectively.

Step-by-step solution

Write the balanced chemical equations

We have two possible reactions occurring in the cylinder: 1. Graphite (C) reacts with oxygen (O\(_2\)) to form carbon monoxide (CO): \(C + \frac{1}{2}O_2 \rightarrow CO\) 2. Graphite (C) reacts with oxygen (O\(_2\)) to form carbon dioxide (CO\(_2\)): \(C + O_2 \rightarrow CO_2\)

Set up the relationship between moles of CO and CO\(_2\)

Let x moles of CO be formed in the reaction, then we have 5 - x moles of CO\(_2\) formed. (5 moles of Carbon are used up)

Determine the total moles and moles of O\(_2\) left

Since we started with 5 moles of Oxygen, oxygen moles consumed to produce CO = \(\frac{1}{2}\) * x moles, and consumed to produce CO\(_2\) = (5 - x) moles. So, moles of O\(_2\) left = 5 - \(\frac{1}{2}\)x - 5 + x = \(\frac{1}{2}\)x moles. Total moles after reaction = x + 5 - x + \(\frac{1}{2}\)x = 5 + \(\frac{1}{2}\)x.

Calculate the change in pressure

We will use the ideal gas law equation, \(PV=nRT\). Since the temperature and the volume remain constant, the initial and final pressure ratios will relate to their respective mole ratios. The increase in pressure is 17%: \(\frac{P_{final}}{P_{initial}} = 1 + 0.17 = 1.17\) We can write this relation in terms of moles: \(\frac{5 + \frac{1}{2}x}{5} = 1.17\) (Because initial moles were 5)

Solve for x

Solve the equation from step 4 for x: \(\frac{1}{2}x = 5(0.17)\) x = 10(0.17) x = 1.7 moles

Calculate mole fractions

Now that we have the moles of CO (1.7), we can find the moles of CO\(_2\) formed as well (5 - 1.7 = 3.3). And we also find that \(\frac{1}{2}\) * 1.7 = 0.85 moles of O\(_2\) are left. Mole fraction of CO = \(\frac{1.7}{1.7 + 3.3 + 0.85} = \frac{1.7}{5.85}\) ≈ 0.291 Mole fraction of CO\(_2\) = \(\frac{3.3}{5.85}\) ≈ 0.564 Mole fraction of O\(_2\) = \(\frac{0.85}{5.85}\) ≈ 0.145 After the combustion reaction, the mole fractions of CO, CO\(_2\), and O\(_2\) in the final mixture are approximately 0.291, 0.564, and 0.145, respectively.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It’s expressed as \(PV = nRT\), where \(P\) stands for pressure, \(V\) for volume, \(n\) for the number of moles, \(R\) is the ideal gas constant, and \(T\) represents the temperature in kelvins.
Understanding this law allows us to predict how changes in one property of a gas affect the others, provided the gas behaves ideally. Ideal behavior assumes that gas molecules do not interact and occupy no space; real gases approximate this under low pressure and high temperature.
In our problem, we used the Ideal Gas Law to relate the change in pressure to the change in moles of gases. The pressure increased by 17% because the number of moles in the cylinder increased after the reactions.

Balanced Chemical Equations

Balanced chemical equations are essential in stoichiometry as they ensure that atoms are conserved in a reaction. The coefficients in a balanced equation show the ratio of moles needed for reactants and products.
For this problem, it was crucial to balance the equations for the conversion of carbon (graphite) to carbon monoxide (\(CO\)) and carbon dioxide (\(CO_2\)):

• \(C + \frac{1}{2}O_2 \rightarrow CO\)
• \(C + O_2 \rightarrow CO_2\)

These equations help us understand that it requires different amounts of oxygen to produce \(CO\) and \(CO_2\). Balancing these reactions allows us to calculate the moles of leftover reactants and generated products.

Mole Fraction

Mole fraction is a way of expressing the ratio of moles of a particular component to the total moles of a mixture. It is denoted as \(X\), such that the sum of mole fractions in a mixture always equals 1.
In this exercise, once we determined the moles of the products \(CO\) and \(CO_2\), and the remaining \(O_2\), we calculated their specific mole fractions as follows:

• Mole fraction of \(CO\) = \(\frac{1.7}{5.85}\)
• Mole fraction of \(CO_2\) = \(\frac{3.3}{5.85}\)
• Mole fraction of \(O_2\) = \(\frac{0.85}{5.85}\)

Mole fractions help in understanding the composition of the final gaseous mixture formed post-reaction.

Combustion Reaction

Combustion reactions involve the burning of a substance in the presence of oxygen to produce energy, often yielding heat and light. They are generally highly exothermic.
In our problem, graphite and oxygen undergo combustion when ignited, forming mixtures of \(CO\) and \(CO_2\). The balanced equations for these reactions reflect two possible stoichiometric pathways for the carbon: partial oxidation to \(CO\) or complete oxidation to \(CO_2\).
This dual-path process explains why we have a mixture of \(CO\) and \(CO_2\) as products. Understanding combustion reactions requires knowledge of different potential reaction pathways and the energy changes associated with them.