Problem 98 - Chemistry and Chemical Reactivity

Question

A 1.50 L constant volume calorimeter (Figure 5.12 ) contains \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) The partial pressure of \(\mathrm{C}_{3} \mathrm{H}_{8}\) is 0.10 atm and the partial pressure of \(\mathrm{O}_{2}\) is 5.0 atm. The temperature is \(20.0^{\circ} \mathrm{C}\). A reaction occurs between the two compounds, forming \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\ell) .\) The heat from the reaction causes the temperature to rise to \(23.2^{\circ} \mathrm{C}\) (a) Write a balanced chemical equation for the reaction. (b) How many moles of \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})\) are present in the flask initially? (c) What is the mole fraction of \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})\) in the flask before reaction? (d) After the reaction, the flask contains excess oxygen and the products of the reaction, \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\ell) .\) What amount of unreacted \(\mathrm{O}_{2}(\mathrm{g})\) remains? (e) After the reaction, what is the partial pressure exerted by the \(\mathrm{CO}_{2}(\mathrm{g})\) in this system? (f) What is the partial pressure exerted by the excess oxygen remaining after the reaction?

Step-by-Step Solution

Answer

(a) Balanced equation is \(\mathrm{C}_{3}\mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O}\). (b) 0.0062 moles of \(\mathrm{C}_{3}\mathrm{H}_{8}\). (c) Mole fraction is 0.0195. (d) 0.280 moles unreacted \(\mathrm{O}_{2}\). (e) \(\mathrm{CO}_{2}\) partial pressure is 0.3 atm. (f) Excess \(\mathrm{O}_{2}\) partial pressure is 4.54 atm.

Step 1: Write a Balanced Chemical Equation

The reaction between propane (\(\mathrm{C}_{3}\mathrm{H}_{8}\)) and oxygen (\(\mathrm{O}_{2}\)) produces carbon dioxide (\(\mathrm{CO}_{2}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)). The balanced chemical equation is:\[\mathrm{C}_{3}\mathrm{H}_{8} (g) + 5 \mathrm{O}_{2} (g) \rightarrow 3 \mathrm{CO}_{2} (g) + 4 \mathrm{H}_{2}\mathrm{O} (\ell)\]

Step 2: Calculate Moles of Propane

The ideal gas law equation \( PV = nRT \) can be rearranged to find the number of moles \( n \). Given the partial pressure \( P = 0.10 \ \text{atm} \), volume \( V = 1.50 \ \text{L} \), R = 0.0821 \ \text{L·atm/K·mol}\, and temperature \( T = 293.15 \ \text{K} \), we calculate:\[n = \frac{PV}{RT} = \frac{0.10 \times 1.50}{0.0821 \times 293.15} \approx 0.0062 \ \text{moles}\]Thus, there are approximately 0.0062 moles of \(\mathrm{C}_{3} \mathrm{H}_{8}\) initially.

Step 3: Calculate Mole Fraction of Propane

The mole fraction is calculated using the equation \( x_{\mathrm{C}_{3}\mathrm{H}_{8}} = \frac{n_{\mathrm{C}_{3}\mathrm{H}_{8}}}{n_{\mathrm{total}}} \). Find \( n_{\mathrm{O}_{2}} \) using the ideal gas law for \(\mathrm{O}_{2}\):\[n_{\mathrm{O}_{2}} = \frac{5.0 \times 1.50}{0.0821 \times 293.15} \approx 0.311 \ \text{moles}\]The total moles initially is: \[n_{\mathrm{total}} = 0.0062 + 0.311 = 0.3172 \ \text{moles}\]Hence,\[x_{\mathrm{C}_{3}\mathrm{H}_{8}} = \frac{0.0062}{0.3172} \approx 0.0195\]

Step 4: Calculate Unreacted Oxygen

From the balanced equation, 1 mole of \(\mathrm{C}_{3}\mathrm{H}_{8}\) reacts with 5 moles of \(\mathrm{O}_{2}\). Therefore 0.0062 moles of \(\mathrm{C}_{3}\mathrm{H}_{8}\) will react with \(5 \times 0.0062 = 0.031\) moles of \(\mathrm{O}_{2}\). Initially, there are 0.311 moles of \(\mathrm{O}_{2}\), and hence the unreacted amount is:\[\text{Unreacted } \mathrm{O}_{2} = 0.311 - 0.031 = 0.280 \ \text{moles}\]

Step 5: Determine Partial Pressure of CO2

According to the balanced equation, 1 mole of \(\mathrm{C}_{3}\mathrm{H}_{8}\) produces 3 moles of \(\mathrm{CO}_{2}\). Thus, 0.0062 moles of \(\mathrm{C}_{3}\mathrm{H}_{8}\) produce \(3 \times 0.0062 = 0.0186\) moles of \(\mathrm{CO}_{2}\). Use the ideal gas law:\[P = \frac{nRT}{V} = \frac{0.0186 \times 0.0821 \times 296.15}{1.50} \approx 0.3 \ \text{atm}\]The partial pressure exerted by \(\mathrm{CO}_{2}\) is approximately 0.3 atm.

Step 6: Determine Partial Pressure of Excess Oxygen

The unreacted \(\mathrm{O}_{2}\) is 0.280 moles. The partial pressure of \(\mathrm{O}_{2}\) is given by the ideal gas law:\[P = \frac{nRT}{V} = \frac{0.280 \times 0.0821 \times 296.15}{1.50} \approx 4.54 \ \text{atm}\]The partial pressure exerted by the excess \(\mathrm{O}_{2}\) is approximately 4.54 atm.

Key Concepts

Ideal Gas Law

The ideal gas law is a cornerstone concept in chemistry and physics. It is used to determine how gases behave under different conditions of pressure, volume, and temperature. The ideal gas law is represented by the equation \( PV = nRT \). Here's a quick breakdown of each variable:

• \( P \) is the pressure of the gas, often measured in atmospheres (atm).
• \( V \) represents the volume that the gas occupies, usually in liters (L).
• \( n \) stands for the number of moles of the gas.
• \( R \) is the universal gas constant, which has a value of 0.0821 L·atm/K·mol.
• \( T \) is the temperature measured in Kelvin (K).

In our problem, this law helps us find out how many moles of propane, \( \text{C}_3\text{H}_8 \), are present initially. Given that the gas is contained in a 1.50 L calorimeter at a pressure of 0.10 atm and a temperature of 293.15 K, we can easily apply the ideal gas law. This illustrates how practical and useful this equation is in experimental and theoretical scenarios.

Balanced Chemical Equation

Writing a balanced chemical equation is essential to understanding the basics of chemical reactions. A balanced equation ensures that the same number of each type of atom appears on both sides of the reaction. This abides by the law of conservation of mass, which states that matter cannot be created or destroyed.

In the given problem, propane (\( \text{C}_3\text{H}_8 \)) reacts with oxygen (\( \text{O}_2 \)) to form carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)). The balanced equation is:
\[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \]
This equation tells us that one mole of propane reacts with five moles of oxygen. In the process, it produces three moles of carbon dioxide and four moles of water. By balancing the equations, we can accurately deduce how much of each reactant is necessary and what amounts of products will be formed.

Mole Fraction

The mole fraction is a way of expressing the composition of a gas mixture. It is the ratio of the number of moles of one component to the total number of moles in the mixture. This gives us an understanding of the gas's composition by proportion.

The formula for mole fraction \( (x) \) is given by:
\[ x_{\text{component}} = \frac{n_{\text{component}}}{n_{\text{total}}} \]
Using the ideal gas law, we found the moles of \( \text{O}_2 \) and \( \text{C}_3\text{H}_8 \). The total moles in the system before the reaction begins can then be calculated. Once we have the total, the mole fraction of propane can be computed by dividing the moles of propane by the total moles.
In our scenario, the mole fraction of propane is approximately 0.0195, indicating that it constitutes only a small portion of the total gas mixture.

Partial Pressure

Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. It is equivalent to the pressure that the gas would exert if it occupied the entire volume on its own. Understanding partial pressure is crucial in predicting how gases will behave in mixtures.

The partial pressure of a gas can be calculated using the ideal gas law and is directly proportional to its mole fraction:
\[ P_{\text{gas}} = x_{\text{gas}} \times P_{\text{total}} \]
In the post-reaction scenario, the partial pressure of \( \text{CO}_2 \) and the unreacted \( \text{O}_2 \) were estimated. The former is approximately 0.3 atm, and the latter remains at about 4.54 atm.
Such calculations are especially applicable when determining the physical behavior of gases in closed systems, like in a calorimeter or any constrained environment. Understanding these pressures helps in evaluating the efficiency and completion of reactions, among other analytical and practical applications.