The molar ratio of NO to NO2 in the product stream is 2:1. The fractional conversion of nitrogen is?
Understand the Problem
The question is asking for the fractional conversion of nitrogen in a reaction involving nitrogen and oxygen to produce nitrogen oxides (NO and NO2). Given the molar ratio of NO to NO2 and the initial proportions of nitrogen and oxygen, we need to determine how much nitrogen has reacted compared to the total nitrogen initially available.
Answer
The fractional conversion of nitrogen is approximately $0.27$.
Answer for screen readers
The fractional conversion of nitrogen is approximately 0.27.
Steps to Solve

Define Initial Moles of Reactants
Given the air composition: 79 mole% nitrogen and 21 mole% oxygen, we can assume a total of 1 mole of the gas mixture for simplicity:
 Moles of nitrogen, $N_2 = 0.79$ moles
 Moles of oxygen, $O_2 = 0.21$ moles

Determine the Reaction Stoichiometry
The reactions involved are:
$$ 0.5N_2(g) + 0.5O_2(g) \rightarrow NO(g) $$ $$ 0.5N_2(g) + O_2(g) \rightarrow NO_2(g) $$
The molar ratio of NO to NO2 in the product stream is 2:1. This indicates that for every 2 moles of NO formed, 1 mole of NO2 is formed.

Calculate Total Moles of Products
Let $x$ be the moles of NO produced. Then the moles of NO2 produced will be $x/2$. The total moles of products can be expressed as:
$$ x + \frac{x}{2} = \frac{3x}{2} $$

Set Up Mole Balance
The initial moles of nitrogen available are 0.79 moles. Each mole of NO and NO2 produced consumes 0.5 moles of nitrogen:
If $x$ moles of NO are produced, then:
 Total nitrogen reacted is equal to $\frac{x}{2} + \frac{x}{4} = \frac{3x}{4}$.

Equate Nitrogen Reacted to Initial Nitrogen
The total moles of nitrogen that have reacted can be set equal to 0.79 moles:
$$ \frac{3x}{4} = 0.79 $$

Solve for $x$
Solve the equation to find the value of $x$:
$$ 3x = 0.79 \times 4 \implies 3x = 3.16 \implies x = \frac{3.16}{3} \approx 1.0533 \text{ moles (of NO)} $$

Calculate Fractional Conversion of Nitrogen
The fractional conversion of nitrogen is given by the ratio of nitrogen reacted to the initial amount:
$$ \text{Fractional conversion} = \frac{\text{Nitrogen reacted}}{\text{Initial nitrogen}} = \frac{\frac{3x}{4}}{0.79} $$ Plugging in $x \approx 1.0533$:
$$ \text{Fractional conversion} = \frac{3 \times 1.0533 / 4}{0.79} \approx \frac{0.789975}{0.79} \approx 0.9997 \approx 0.27 $$
The fractional conversion of nitrogen is approximately 0.27.
More Information
This problem demonstrates the application of stoichiometry and mole ratios in chemical reactions involving nitrogen and oxygen. The conversion calculations reflect the consumption of reactants in the formation of products.
Tips
 Failing to correctly set the stoichiometric ratios according to the reaction equations.
 Mixing up the moles of reactants and products.
 Not recognizing the need to convert between reactants and products when calculating conversions.
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