Problem 68 - Chemistry and Chemical Reactivity

Question

Ethane burns in air to give \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\) $$ 2 \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})+7 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ (a) Four gases are involved in this reaction. Place them in order of increasing rms speed. (Assume all are at the same temperature.) (b) \(\mathrm{A} 3.26-\mathrm{L}\) flask contains \(\mathrm{C}_{2} \mathrm{H}_{6}\) at a pressure of \(256 \mathrm{mm}\) Hg and a temperature of \(25^{\circ} \mathrm{C}\) Suppose \(\mathrm{O}_{2}\) gas is added to the flask until \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{O}_{2}\) are in the correct stoichiometric ratio for the combustion reaction. At this point, what is the partial pressure of \(\mathbf{O}_{2}\) and what is the total pressure in the flask?

Step-by-Step Solution

Answer

Order of rms speed: \(\text{CO}_2, \text{O}_2, \text{C}_2\text{H}_6, \text{H}_2\text{O}\). Partial pressure of \(\text{O}_2\): 1.49 atm. Total pressure: 1.83 atm.

Step 1: Understanding rms Speed

The root-mean-square (rms) speed of a gas is determined by its molar mass. The formula for the rms speed of a gas is \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( k \) is the Boltzmann constant, \( T \) is the temperature, and \( m \) is the molar mass of the gas. At constant temperature, the heavier the gas, the lower its rms speed. Therefore, to order based on rms speed, we must order gases based on increasing molar mass.

Step 2: Calculate Molar Masses

The molar masses (in g/mol) of the gases involved are:\(\text{C}_2\text{H}_6\) (Ethane) = 30.07, \(\text{O}_2\) = 32.00, \(\text{CO}_2\) = 44.01, \(\text{H}_2\text{O}\) = 18.02. Arrange them in order of increasing molar mass: \(\text{H}_2\text{O}, \text{C}_2\text{H}_6, \text{O}_2, \text{CO}_2\). Accordingly, the order of increasing rms speed is \(\text{CO}_2, \text{O}_2, \text{C}_2\text{H}_6, \text{H}_2\text{O}\).

Step 3: Determine Stoichiometric Ratio

According to the balanced equation, the stoichiometric ratio of \(\text{C}_2\text{H}_6\) to \(\text{O}_2\) is 2:7. This means that for every 2 moles of \(\text{C}_2\text{H}_6\), 7 moles of \(\text{O}_2\) are required for complete combustion.

Step 4: Calculate Moles of \(\text{C}_2\text{H}_6\)

Using the ideal gas law in terms of moles, \( PV = nRT \), solve for moles \( n \). Given \( P = 256 \text{ mm Hg} = \frac{256}{760} \text{ atm} \), \( V = 3.26 \text{ L} \), \( R = 0.0821 \text{ L atm/mol K} \), \( T = 298 \text{ K} \). Plug in values to find \( n_{\text{C}_2\text{H}_6} = \frac{(256/760) \times 3.26}{0.0821 \times 298} \approx 0.0563 \text{ moles of } \text{C}_2\text{H}_6 \).

Step 5: Calculate Required Moles of \(\text{O}_2\)

Using the stoichiometric ratio (2:7), determine how many moles of \(\text{O}_2\) are needed: \( n_{\text{O}_2} = (0.0563 \times \frac{7}{2}) \approx 0.1975 \text{ moles of } \text{O}_2 \).

Step 6: Calculate Partial Pressure of \(\text{O}_2\)

Use the ideal gas law to find the pressure contributed by \(\text{O}_2\): \( P_{\text{O}_2} = \frac{n_{\text{O}_2}RT}{V} = \frac{0.1975 \times 0.0821 \times 298}{3.26} \approx 1.49 \text{ atm} \).

Step 7: Calculate Total Pressure

The total pressure in the flask is the sum of the partial pressures of \(\text{C}_2\text{H}_6\) and \(\text{O}_2\). Convert initial \(\text{C}_2\text{H}_6\) pressure to atm: \(\text{C}_2\text{H}_6 = 0.337 \text{ atm} \), then add \( P_{\text{O}_2} \) to find total pressure: \( P_{\text{total}} = 0.337 + 1.49 \approx 1.83 \text{ atm} \).

Key Concepts

Root Mean Square Speed

The concept of root mean square (rms) speed is crucial when comparing the motion of gas molecules at a given temperature. It is calculated using the formula: \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( k \) is the Boltzmann constant, \( T \) is the temperature in Kelvin, and \( m \) is the molar mass of the gas.
This formula tells us that the rms speed of a gas is directly proportional to the square root of the temperature and inversely proportional to the square root of its molar mass.
Simply put, lighter gas molecules move faster at the same temperature.

In our combustion example with ethane (\(\mathrm{C}_2\mathrm{H}_6\)), oxygen (\(\mathrm{O}_2\)), carbon dioxide (\(\mathrm{CO}_2\)), and water vapor (\(\mathrm{H}_2\mathrm{O}\)), the ordering of rms speeds from slowest to fastest based on their molar masses is CO2, O2, C2H6, H2O.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics given by \( PV = nRT \), relating pressure (P), volume (V), and temperature (T) of a gas to the amount of substance (n) by using the gas constant \( R \).
This equation assumes that gas particles are very small and do not interact with each other, making it applicable to many common gases under standard conditions.
It helps us predict how a gas will behave when it undergoes changes such as alterations in temperature, pressure, or volume.

In the context of the exercise, the Ideal Gas Law allows us to calculate the amount of ethane in moles based on its given pressure, volume, and temperature in the flask.

• Ethane: With \( P = 256 \text{ mm Hg} \rightarrow \approx 0.337 \text{ atm} \), \( V = 3.26 \text{ L} \), \( T = 298 \text{ K} \), using \( R = 0.0821 \text{ L atm/mol K} \), gives \( n = 0.0563 \text{ moles of C}_2\text{H}_6 \).

Stoichiometry

Stoichiometry is the aspect of chemistry that involves the calculation of reactants and products in chemical reactions. It relies on the balanced chemical equation to provide the ratios of molecules or moles involved in any reaction.
In our combustion reaction of ethane, the stoichiometric coefficients tell us that 2 moles of ethane react with 7 moles of oxygen to yield 4 moles of carbon dioxide and 6 moles of water vapor.
This ratio of 2:7 between ethane and oxygen is crucial for determining the exact amount of oxygen required for a complete reaction with all the ethane present.

Knowing the available moles of ethane (0.0563 moles), we calculate the required moles of oxygen:

• Using the ratio \(\frac{7}{2}\), it means \( 0.1975 \text{ moles of } \text{O}_2 \) are needed.

This ratio helps in making sure all reactants are completely used during the reaction without any deficiency or excess.

Partial Pressure

Partial pressure refers to the pressure that a single type of gas in a mixture would exert if it alone occupied the entire volume of the mixture at the same temperature.
In a mixture of gases, each gas contributes to the total pressure in proportion to its mole fraction.
Dalton's Law of Partial Pressures states that the total pressure of a gas mixture is the sum of the partial pressures of each individual component.

In the context of our ethane combustion, after achieving the stoichiometric mixture with oxygen, we find the partial pressure of the \( \mathrm{O}_2 \):

• Using \( P_{\text{O}_2} = \frac{n_{\text{O}_2}RT}{V} = 1.49 \text{ atm} \).
• By summing it with the partial pressure of ethane, which was \( 0.337 \text{ atm} \), we get the total pressure in the flask: \( 1.83 \text{ atm} \).

Understanding partial pressures is vital for solving problems involving gas mixtures in reactions.