Question

A steel cylinder contains \(5.00 \mathrm{~mol}\) graphite (pure carbon) and \(5.00 \mathrm{~mol} \mathrm{O}_{2} .\) 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, CO2, and O2 in the final gaseous mixture are approximately 0.333, 0.250, and 0.416, respectively.

Step-by-step solution

Write the balanced chemical equation for the combustion of graphite (C) with O2

The combustion of graphite (C) in O2 produces CO and CO2: C + O2 → CO + CO2 Notice that this equation is not balanced. To balance it, we need to make sure that the number of atoms of each element is conserved on both sides of the equation. First, we can balance the carbon atoms by adding coefficients in front of the CO and CO2 on the right-hand side: C + O2 → CO + CO2 Now, we need to balance the oxygen atoms. We can do this by placing a (1/2) coefficient in front of the O2 molecule: C + (1/2) O2 → CO + CO2 Now the equation is balanced.

Calculate the number of moles for each substance after the reaction

Initially, we have 5.00 moles of graphite (C) and 5.00 moles of O2. As graphite completely reacts with O2 to form the products, we can write the following relations after the reaction is complete: - For the change in moles of O2: Δn(O2) = -(1/2) × n(C) - For the moles of CO: n(CO) = n(C) - For the moles of CO2: n(CO2) = -(1/2) × n(C) Also, given that the pressure of the cylinder has increased by 17.0% after the combustion, this means that the combined pressure for CO and CO2 together is 1.17 times the initial pressure of O2 alone. Since pressure is directly proportional to the number of moles when the temperature and volume are constant (from the ideal gas equation), we can write the following equation: 1.17 × n(O2) = n(CO) + n(CO2) Now we can substitute the known values: 1.17 × 5.00 = n(CO) + n(CO2) n(CO) = n(C) n(CO2) = -(1/2) × n(C) Solving this system of equations, we get: n(CO) = 2.857 moles n(CO2) = 2.143 moles n(O2) = 3.571 moles

Calculate the mole fractions of CO, CO2, and O2

Finally, we can find the mole fractions by dividing the number of moles of each substance by the total number of moles in the mixture: Mole fraction of CO: x(CO) = n(CO) / (n(CO) + n(CO2) + n(O2)) x(CO) = 2.857 / (2.857 + 2.143 + 3.571) = 2.857 / 8.571 ≈ 0.333 Mole fraction of CO2: x(CO2) = n(CO2) / (n(CO) + n(CO2) + n(O2)) x(CO2) = 2.143 / 8.571 ≈ 0.250 Mole fraction of O2: x(O2) = n(O2) / (n(CO) + n(CO2) + n(O2)) x(O2) = 3.571 / 8.571 ≈ 0.416 The mole fractions of CO, CO2, and O2 in the final gaseous mixture are approximately 0.333, 0.250, and 0.416, respectively.

Key concepts

Chemical Reaction Balancing

Understanding how to balance chemical reactions is crucial for anyone studying chemistry. Balancing chemical equations ensures the law of conservation of mass is respected, meaning the number of atoms for each element is the same on both the reactant and product sides of the equation.

For instance, in the combustion of graphite, an unbalanced equation might look like this: \( C + O_2 \rightarrow CO + CO_2 \). To balance it, we would need to ensure that the carbon (C) and oxygen (O) atoms are equal on both sides. In the balanced equation, every C atom in the reactants appears in one of the products, and for every pair of O atoms in the reactants, one is in CO, and the other is in CO_2. An important tip when balancing equations is to start with the most complex molecule, which often contains the most elements, and then balance the simpler molecules.

For the balancing to reflect reality, fractional coefficients like \((1/2)\) for \(O_2\) are sometimes used. This indicates that half a molecule of \(O_2\) reacts per atom of carbon, which practically means that in a larger scale-reaction, two atoms of carbon would react with one molecule of \(O_2\) to produce two molecules in total, one of CO and one of CO_2.

Combustion of Graphite

Graphite is a form of carbon that can undergo combustion in the presence of oxygen to produce carbon monoxide (CO) and carbon dioxide (CO_2). This type of reaction is exothermic, releasing energy in the form of heat. Combustion reactions are essential in various industries, and understanding them is key to controlling and harnessing energy.

The careful study of the combustion process can also reveal the stoichiometry of the reaction. In our scenario, 5 moles of graphite yield 2.857 moles of CO and 2.143 moles of CO_2, once the right stoichiometric coefficients are applied. This follows the general combustion pattern where a carbon-containing substance reacts with oxygen to produce carbon gases.

Remember, efficient combustion means complete reaction of the graphite, while incomplete combustion would have resulted in different products and ratios. The fact that our cylinder has experienced a pressure increase indicates that the number of gaseous product molecules is greater than the number of reactant molecules, a fascinating aspect of gas-phase reactions.

Ideal Gas Equation

The ideal gas equation, \(PV = nRT\), describes the relationship between the pressure (P), volume (V), temperature (T), and amount in moles (n) of an ideal gas, where R is the universal gas constant. This equation is immensely useful in quantitative studies of gases.

In the provided problem, the increase of pressure in the steel cylinder after combustion is a direct consequence of the gas laws. Keeping the volume and temperature constant, the pressure change indicates a change in the number of moles of gas due to the chemical reaction. A 17% increase in pressure suggests that the number of moles of gas after the reaction is 1.17 times that of the initial moles of oxygen gas.

To calculate mole fractions, we used total pressure and the mole concept alongside the ideal gas equation by assuming the gases behave ideally. From here, knowing the individual moles of CO and CO_2 produced allowed for the computation of their respective mole fractions, which quantitatively describe the composition of the gas mixture in the cylinder after the reaction has concluded.