Problem 83 - Chemistry and Chemical Reactivity

Question

Nitrogen monoxide reacts with oxygen to give nitrogen dioxide: $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) $$ (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 pres. sure of \(150 \mathrm{mm} \mathrm{Hg}\) and \(\mathrm{O}_{2}\) was added in the correct stoichiometric amount?

Step-by-Step Solution

Answer

(a) \(\mathrm{NO}_2\) < \(\mathrm{O}_2\) < \(\mathrm{NO}\), (b) 75 mm Hg, (c) 150 mm Hg.

Step 1: Understanding rms speed

The root mean square (rms) speed of a gas is determined by the formula \( u_{rms} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas in kg/mol. Since \( T \) is constant at 298 K, the rms speed is inversely proportional to the square root of the molar mass. Thus, gases with lower molar mass will have higher rms speeds.

Step 2: Determining molar masses

The molar masses are as follows: \(\mathrm{NO} = 14 + 16 = 30 \ \mathrm{g/mol}\), \(\mathrm{O}_2 = 2 \times 16 = 32 \ \mathrm{g/mol}\), and \(\mathrm{NO}_2 = 14 + 2 \times 16 = 46 \ \mathrm{g/mol}\).

Step 3: Order gases by decreasing rms speed

According to their molar masses, the order of increasing rms speed is: \(\mathrm{NO}_2\) (slowest) < \(\mathrm{O}_2\) < \(\mathrm{NO}\) (fastest).

Step 4: Calculate partial pressure of \(\mathrm{O}_2\)

From the balanced reaction, the stoichiometric ratio of \(\mathrm{NO}\) to \(\mathrm{O}_2\) is 2:1. Given \(\mathrm{NO}\) has a partial pressure of 150 mm Hg, the partial pressure of \(\mathrm{O}_2\) will be half of this, being 75 mm Hg. This follows from the stoichiometric ratio.

Step 5: Find final pressure of \(\mathrm{NO}_2\)

For every 2 mol of \(\mathrm{NO}\), 2 mol of \(\mathrm{NO}_2\) are formed without changing the number of total gas moles. Since we're using 150 mm Hg \(NO\) and the correct amount of \(\mathrm{O}_2\) (75 mm Hg), all will convert to \(\mathrm{NO}_2\). The total pressure of \(\mathrm{NO}_2\) will be the same as the initial \(\mathrm{NO}\), which is 150 mm Hg.

Key Concepts

Root Mean Square Speed

The root mean square (rms) speed is a concept in chemistry that helps to describe how fast gas molecules, on average, are moving. It's calculated using the formula \[ u_{rms} = \sqrt{\frac{3RT}{M}} \] where:

• \( u_{rms} \) is the root mean square speed,
• \( R \) is the universal gas constant,
• \( T \) is the temperature in Kelvin,
• \( M \) is the molar mass of the gas in kilograms per mole (kg/mol).

One crucial point about this formula is that it's rooted in the principles of the kinetic molecular theory.
In practice, because temperature \( T \) is constant in many problems, gases with a lower molar mass \( M \) will typically have a higher rms speed, making lighter gases move faster than heavier ones. In the context of the chemical reaction given, at 298 K:

• Nitrogen dioxide \(NO_2\) has the highest molar mass of 46 g/mol,
• Oxygen \(O_2\) follows at 32 g/mol,
• and finally, nitrogen monoxide \(NO\) has the lowest molar mass of 30 g/mol.

Therefore, when arranging these gases by increasing rms speed, the order is \( NO_2 \) (slowest) < \( O_2 \) < \( NO \) (fastest).

Partial Pressure Calculation

In chemical reactions involving gases, calculating partial pressures becomes important, especially when stoichiometry is considered. Partial pressure is the pressure that each gas in a mixture would exert if it alone occupied the entire volume of the mixture.
For a balanced chemical equation such as this one:\[ 2 \mathrm{NO} + \mathrm{O}_2 \rightarrow 2 \mathrm{NO}_2 \]the stoichiometric coefficients indicate the molar ratio of the reactants and products.
Here, we see the ratio of \(\mathrm{NO}\) to \(\mathrm{O}_2\) is 2:1, meaning two moles of \(\mathrm{NO}\) react with one mole of \(\mathrm{O}_2\).
Given that the partial pressure of \(\mathrm{NO}\) is 150 mm Hg, the partial pressure of \(\mathrm{O}_2\) can be calculated using this stoichiometric relationship.
Since the ratio is 2:1, the partial pressure of \(\mathrm{O}_2\) will be half that of \(\mathrm{NO}\), resulting in a value of 75 mm Hg.
This basic calculation aligns with the idea of maintaining proportional relationships, as dictated by chemical equations in gaseous forms.

Kinetic Molecular Theory

The kinetic molecular theory (KMT) is central to understanding gas behavior, providing a conceptual framework that explains the motion of particles in gases. Fundamentally, it outlines several key postulates:

• Gases consist of numerous tiny particles that are in constant random motion.
• The volume of the gas particles themselves is negligible compared to the total volume gas occupies.
• There are no forces of attraction or repulsion between the gas particles.
• Particle collisions are perfectly elastic, meaning no energy is lost during collisions.
• The average kinetic energy of gas particles is proportional to the temperature in Kelvin.

In applying KMT to the problem at hand, the theory supports understanding concepts like rms speed where temperature directly impacts gas particle speeds.
From KMT, we infer that the same temperature will mean all gases, despite differences in mass, have particles that, on average, possess similar kinetic energies. This helps explain why lighter gases tend to move faster, capturing the essence of the exercise. Additionally, the theory underlies the relationships in partial pressure calculations by asserting that temperature, pressure, volume, and number of particles are all interrelated properties that predict gas behavior. Every adjustment in these parameters relates back to the inherent kinetic motions of the gas molecules, thus holistically tying together stoichiometry and gaseous behavior.