Problem 89 - Chemistry and Chemical Reactivity

Question

Nitrogen monoxide reacts with oxygen to give nitrogen dioxide: $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) $$ (a) Place the three gases in order of increasing rms speed at \(298 \mathrm{K}\) (b) If you mix \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) in the correct stoichiometric ratio and NO has a partial pressure of \(150 \mathrm{mm} \mathrm{Hg},\) what is the partial pressure of \(\mathrm{O}_{2} ?\) (c) After reaction between \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) is complete, what is the pressure of \(\mathrm{NO}_{2}\) if the NO originally had a pressure of \(150 \mathrm{mm} \mathrm{Hg}\) and \(\mathrm{O}_{2}\) was added in the correct stoichiometric amount?

Step-by-Step Solution

Answer

(a) NO_2, O_2, NO; (b) 75 mm Hg; (c) 150 mm Hg.

Step 1: Understand Root Mean Square (RMS) Speed

The RMS speed of a gas molecule at a given temperature can be found using the formula: \( \text{RMS speed} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas. This implies that RMS speed is inversely proportional to the square root of the molar mass.

Step 2: Determine Molar Mass of Each Gas

Calculate the molar masses: NO has a molar mass of \( 30 \ g/mol \), \( O_2 \) has a molar mass of \( 32 \ g/mol \), and \( NO_2 \) has a molar mass of \( 46 \ g/mol \).

Step 3: Order Gases by Increasing RMS Speed

Using Step 1, the gas with the highest molar mass has the lowest RMS speed. Therefore, the order of increasing RMS speed is \( NO_2 < O_2 < NO \).

Step 4: Determine Oxygen Partial Pressure

From the stoichiometry of the balanced equation, 2 moles of NO react with 1 mole of \( O_2 \). Thus, the partial pressure of \( O_2 \) must be half that of \( NO \), which is \( 75 \ mm Hg \).

Step 5: Calculate Final Pressure of \( NO_2 \)

Assuming the reaction goes to completion, all the \( NO \) and \( O_2 \) convert to \( NO_2 \). Since 2 moles of \( NO \) produce 2 moles of \( NO_2 \), the initial partial pressures of NO will result entirely in \( NO_2 \) pressure. Thus, \( NO_2 \) will have a pressure of \( 150 \ mm Hg + 0 \ mm Hg \) (as the oxygen was added just enough according to stoichiometry). Therefore, \( P_{NO_2} = 150 \ mm Hg \).

Key Concepts

Gas Laws

Gas laws are essential in understanding how gases behave under various conditions. Three key gas laws are Boyle's Law, Charles's Law, and Avogadro's Law, which describe relationships between volume, pressure, and temperature of gases.
Boyle's Law explains that the pressure of a gas is inversely proportional to its volume, provided the temperature and number of moles remain constant.

• This means as the volume increases, the pressure decreases and vice versa.
• The equation used is: \(P_1V_1 = P_2V_2\).

Charles's Law states that the volume of a gas is directly proportional to its Kelvin temperature when the pressure is constant.

• This implies as the temperature increases, so does the volume.
• The formula is: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\).

Finally, Avogadro's Law shows that the volume of a gas is directly proportional to the number of moles, given constant pressure and temperature.

• This can be represented as \(V_1/n_1 = V_2/n_2\).

Together, these laws can be combined in the ideal gas law: \(PV = nRT\), where \(R\) is the gas constant. Understanding these relationships helps in predicting the behavior of gases during chemical reactions under different conditions.

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It allows us to predict how much of a substance is needed or produced in a reaction. One fundamental aspect of stoichiometry is understanding and using the balanced chemical equation. This indicates the mole ratio of reactants and products.
In the given reaction, \(2 \ ext{NO}(g) + \text{O}_2(g) \rightarrow 2 \ ext{NO}_2(g)\), stoichiometry tells us:

• 2 moles of \(NO\) react with 1 mole of \(O_2\).
• If 1 mole of \(O_2\) is used, it will completely react with 2 moles of \(NO\) to produce 2 moles of \(NO_2\).

This process can also be used to calculate the necessary amounts of reactants needed for a reaction to proceed. For example, if we start with a certain pressure of \(NO\), we can use stoichiometry to find the required pressure of \(O_2\). This ensures the reactants are in the correct proportions, leading to a complete reaction and accurate product amounts. Understanding stoichiometry is crucial for efficiency and cost-effectiveness in chemical reactions.

RMS Speed

Root Mean Square (RMS) speed is the measure of the speed of particles in a gas. It is related to the average kinetic energy of gas particles. RMS speed can be calculated using the formula: \[ \text{RMS speed} = \sqrt{\frac{3RT}{M}} \]where:

• \(R\) is the universal gas constant, \(8.314 \ J/(mol\cdot K)\).
• \(T\) is the absolute temperature in Kelvin.
• \(M\) is the molar mass of the gas in kilograms per mole.

This equation shows that the speed of gas molecules increases with temperature and decreases with increased molar mass.
For instance, in the given exercise:

• \(NO\) has the lowest molar mass (30 g/mol) and hence the highest RMS speed.
• \(O_2\) follows with a slighter greater molar mass.
• \(NO_2\) has the highest molar mass (46 g/mol) and thus the lowest RMS speed.

Understanding RMS speed is crucial in calculating and predicting the behavior of gases in reactions by analyzing their molecular speeds at different temperatures.

Partial Pressure

Partial pressure is a concept that helps us understand the contribution of each gas in a mixture to the total pressure. According to Dalton's Law, in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of individual gases.

• This can be mathematically represented as \(P_{total} = P_1 + P_2 + \ldots + P_n\).

To find the partial pressure of a gas, you need its mole fraction and the total pressure. The formula is: \[ P_i = \chi_i \times P_{total} \]where \(P_i\) is the partial pressure, \(\chi_i\) is the mole fraction, and \(P_{total}\) is the total pressure.
In the exercise, stoichiometry helps determine that since 2 moles of \(NO\) react with 1 mole of \(O_2\), if \(NO\) has a partial pressure of 150 mm Hg, \(O_2\)'s partial pressure must be half, 75 mm Hg. Understanding partial pressures is significant in predicting how gases will behave when mixed, ensuring proper conditions for reactions.