The molar ratio of NO to NO2 in the product stream is 2:1. The fractional conversion of nitrogen is?

Steps to Solve

1. 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

2. 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.

3. 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:

   $$ x + \frac{x}{2} = \frac{3x}{2} $$

4. 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 $\frac{x}{2} + \frac{x}{4} = \frac{3x}{4}$.

5. Equate Nitrogen Reacted to Initial Nitrogen

   The total moles of nitrogen that have reacted can be set equal to 0.79 moles:

   $$ \frac{3x}{4} = 0.79 $$

6. Solve for $x$

   Solve the equation to find the value of $x$:

   $$ 3x = 0.79 \times 4 \implies 3x = 3.16 \implies x = \frac{3.16}{3} \approx 1.0533 \text{ moles (of NO)} $$

7. 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 $x \approx 1.0533$:

   $$ \text{Fractional conversion} = \frac{3 \times 1.0533 / 4}{0.79} \approx \frac{0.789975}{0.79} \approx 0.9997 \approx 0.27 $$