Question

Consider the reaction \( \text{N}_2(g) + 3\text{H_2(g) \rightarrow 2\text{NH}_3(g) \) in a continuous flow reactor under steady-state conditions. The component flow rates at the reactor inlet are: \[ F_{\text{N}_2}^0 = 100 \, \text{mol/s}, \quad F_{\text{H}_2}^0 = 300 \, \text{mol/s}, \quad F_{\text{inert}}^0 = 1 \, \text{mol/s}. \] If the fractional conversion of \( \text{H}_2 \) is \( 0.60 \), the outlet flow rate of \( \text{N}_2 \), in mol/s, rounded off to the nearest integer, is \_\_\_\_\_\_\_.}

Hint:

For steady-state reactor problems, use the fractional conversion and stoichiometric ratios to calculate the moles of reactants and products.

Answer 1

Step 1: Determine the moles of \( \text{H}_2 \) reacted. The fractional conversion of \( \text{H}_2 \) is given as \( 0.60 \). The moles of \( \text{H}_2 \) reacted are: \[ \text{Moles of \( \text{H}_2 \) reacted} = F_{\text{H}_2}^0 \cdot 0.60 = 300 \cdot 0.60 = 180 \, \text{mol/s}. \] Step 2: Use the stoichiometry of the reaction. From the stoichiometry \( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \): \[ \text{Moles of \( \text{N}_2 \) reacted} = \frac{1}{3} \cdot \text{Moles of \( \text{H}_2 \) reacted} = \frac{1}{3} \cdot 180 = 60 \, \text{mol/s}. \] Step 3: Calculate the outlet flow rate of \( \text{N}_2 \). The inlet flow rate of \( \text{N}_2 \) is \( F_{\text{N}_2}^0 = 100 \, \text{mol/s} \). The outlet flow rate is: \[ F_{\text{N}_2,\text{out}} = F_{\text{N}_2}^0 - \text{Moles of \( \text{N}_2 \) reacted} = 100 - 60 = 40 \, \text{mol/s}. \] Step 4: Conclusion. The outlet flow rate of \( \text{N}_2 \) is \( 40 \, \text{mol/s} \).