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

In the final gaseous mixture, the mole fractions of CO, CO2, and O2 are 0.1667, 0.6667, and 0.1667, respectively.

Step-by-step solution

Write the balanced chemical equations for the formation of CO and CO2

In this step, we need to write balanced chemical equations for the formation of both CO and CO2 from the reaction of graphite (C) and oxygen (O2). The balanced reactions are: 1. Formation of CO: \(C + \frac{1}{2}O_2 \rightarrow CO\) 2. Formation of CO2: \(C + O_2 \rightarrow CO_2\)

Determine the initial pressure

Before the combustion, there were 5 moles of graphite (C) and 5 moles of oxygen (O2). Since graphite is a solid and does not contribute to the cylinder's pressure, only the 5 moles of O2 affect the pressure. Let's denote the initial pressure of the cylinder as \(P_{initial}\). Since the pressure of the cylinder increased by 17% after the mixture ignited, we can write the final pressure as: \(P_{final} = P_{initial} + 0.17P_{initial} = 1.17P_{initial}\)

Assign variables for reacting graphite moles with O2

Let's denote the moles of graphite that react with O2 to form CO as x, and the moles of graphite that react with O2 to form CO2 as y. The sum of x and y must equal 5, as that is the total moles of graphite present: \(x + y = 5\)

Calculate moles of O2 consumed

Use the balanced chemical equations to determine the moles of O2 consumed by the reactions: 1. For the formation of CO: \(0.5x \space moles \space of \space O_2\) 2. For the formation of CO2: \(y \space moles \space of \space O_2\) The total moles of O2 consumed is: \(0.5x + y\) Since we started with 5 moles of O2: \(5 - (0.5x + y)\)

Calculate the moles and partial pressures of CO, CO2, and O2 after the reactions

Since the overall pressure increased by 17%, we can relate this to the change in moles of gas through the ideal gas law assuming constant temperature and volume. The initial moles of O2 gas and the final moles of the three gases can be related as follows: \(\frac{1.17P_{initial}}{P_{initial}} = \frac{n_{final}}{n_{initial}}\) \(1.17 = \frac{x + y + (5 - (0.5x + y))}{5}\)

Solve for x and y

Now we have two equations to solve for x and y: 1. \(x + y = 5\) 2. \(1.17 = \frac{x + y + (5 - (0.5x + y))}{5}\) After solving this system of equations, we find that: \(x = 1\) \(y = 4\)

Calculate the mole fractions of CO, CO2, and O2

Now that we have the moles of graphite reacting to form CO and CO2, we can calculate the mole fractions for each gas. Remember that the remaining moles of O2 can be calculated with: Remaining O2 moles = \(5 - (0.5x + y) = 5 - (0.5(1) + 4) = 1\) The total moles of gas present at the end of the reaction are: Total moles = \(x + y + 1 = 1 + 4 + 1 = 6\) The mole fractions are: Mole Fraction of CO: \(\frac{x}{Total \space moles} = \frac{1}{6} = 0.1667\) Mole Fraction of CO2: \(\frac{y}{Total \space moles} = \frac{4}{6} = 0.6667\) Mole Fraction of O2: \(\frac{1}{Total \space moles} = \frac{1}{6} = 0.1667\) In the final gaseous mixture, the mole fractions of CO, CO2, and O2 are 0.1667, 0.6667, and 0.1667, respectively.

Key concepts

Chemical Equations

Understanding chemical equations is crucial for students aiming to master mole fraction calculations and related chemistry problems. A chemical equation represents a chemical reaction by displaying the reactants on the left side and the products on the right, with an arrow pointing from the reactants to the products to symbolize the direction of the transformation. Balancing a chemical equation is akin to ensuring a recipe has the correct proportions of ingredients. It's essential to confirm that the number of atoms for each element is the same on both sides of the equation to comply with the law of conservation of mass.

For instance, in the provided exercise, the formation of CO and CO₂ from graphite (C) and oxygen (O₂) requires creating two balanced chemical equations. Getting these equations right is the first step before you can even think about calculating the mole fractions in the resulting mixture. This process encapsulates the foundational concept of the stoichiometry of a reaction, indicating how much of each reactant goes into making the products.

Stoichiometry

Stoichiometry is the area of chemistry that concerns the quantitative relationships between reactants and products in a chemical reaction. It is grounded in the stoichiometric coefficients found in a balanced chemical equation, which tell us the ratio by which reactants combine to form products. When you're calculating mole fractions, as in the textbook exercise, stoichiometry helps you understand how many moles of each reactant are needed to produce a specific amount of products.

In the solution to the exercise, stoichiometry comes into play when determining the moles of O₂ that react to form CO and CO₂. One mole of graphite requires half a mole of O₂ to form CO and a whole mole to form CO₂. The balanced equations reflect these proportions. With these ratios, we can deduce the amounts of graphite (C) that react in each case, as well as the remaining O₂, which is imperative for subsequent mole fraction calculations.

Ideal Gas Law

The ideal gas law is an equation of state for a hypothetical ideal gas. It describes how pressure (P), volume (V), temperature (T), and moles of gas (n) are interconnected through the law expressed as PV = nRT, with R being the universal gas constant. For the textbook problem, the ideal gas law plays a critical role when we assume the volume and temperature of the gas mixture remain unchanged before and after the reaction. This assumption allows us to link the change in pressure within the cylinder to the change in the number of moles of gas.

The ideal gas law underpins the understanding that if temperature and volume are constant, any change in pressure is directly proportional to a change in the number of moles of gas. The 17% increase in pressure tells us there's been an increase in the number of gas moles post-reaction, which is crucial for finding the values of x and y – the moles of graphite that reacted. This helps further the calculation of mole fractions, showcasing how the ideal gas law supports solving complex chemistry problems by connecting physical properties to chemical quantities.