Question

The gas phase reaction \( \text{A} \rightarrow \text{B} + \text{C} \) is carried out in an ideal PFR achieving 40% conversion of A. The feed has 70 mole % A and 30% inerts. The inlet temperature is 300 K and the outlet temperature is 400 K. The ratio of the outlet to inlet molar concentration of A (assuming ideal gas mixture and uniform pressure) is:

• A. 0.60
• B. 0.30
• C. 0.40
• D. 0.35

Hint:

Concentration correction formula for gas systems: \[ \frac{C_A}{C_{A0}} = \frac{1 - X_A}{1 + \varepsilon_A X_A} \cdot \left(\frac{T_0}{T}\right) \] Substitute parameters directly: \[ \frac{C_A}{C_{A0}} = \frac{1 - 0.4}{1 + (0.7 \times 0.4)} \cdot \left(\frac{300}{400}\right) = \frac{0.6}{1.28} \cdot 0.75 = \frac{0.45}{1.28} = 0.35 \]

Answer 1

The Correct Option is D

Concept: For a gas-phase reaction inside a continuous flow reactor, changes in temperature, pressure, or the total number of moles alter the volumetric flow rate of the gas mixture. This variation in volumetric flow rate must be accounted for when calculating the reactant concentration. Using the Ideal Gas Law, the molar concentration of a component $A$ at any point can be related to its molar flow rate ($F_A$) and the volumetric flow rate ($v$): \[ C_A = \frac{F_A}{v} \] The volumetric flow rate ($v$) varies as a function of fractional conversion ($X_A$) and absolute temperature ($T$) according to the relationship: \[ v = v_0 \cdot (1 + \varepsilon_A \cdot X_A) \cdot \left(\frac{T}{T_0}\right) \cdot \left(\frac{P_0}{P}\right) \] Where $\varepsilon_A$ is the fractional change in the total number of moles in the system per mole of reactant $A$ reacted.

Step 1: Calculating the fractional change in moles parameter (\(\varepsilon_A\)).
Let us extract the reaction stoichiometry and feed composition parameters from the problem text: \[ \text{Reaction:} \quad 1\text{A} \rightarrow 1\text{B} + 1\text{C} \] The change in the number of moles for this stoichiometry is: \[ \Delta n = \text{Moles of products} - \text{Moles of reactants} = (1 + 1) - 1 = +1 \] The feed mixture contains:

• Mole fraction of reactant A: \( y_{A0} = 0.70 \)

• Mole fraction of inerts: \( y_{I0} = 0.30 \)
The parameter $\varepsilon_A$ is calculated by multiplying the mole fraction of reactant $A$ in the feed by the stoichiometric change in moles ($\Delta n$): \[ \varepsilon_A = y_{A0} \cdot \Delta n = 0.70 \cdot (+1) = 0.70 \]

Step 2: Setting up the equations for inlet and outlet volumetric flow rates.
The problem states that the system operates at a uniform pressure, meaning $P = P_0$. Let us write the expressions for the volumetric flow rates at the inlet and outlet conditions:

• At the inlet (\( X_A = 0 \), \( T_0 = 300 \, \text{K} \)): \[ v_0 = v_0 \]

• At the outlet (\( X_A = 0.40 \), \( T = 400 \, \text{K} \)): \[ v = v_0 \cdot [1 + \varepsilon_A \cdot X_A] \cdot \left(\frac{T}{T_0}\right) \]
Let us substitute the numerical values into the expression for the outlet volumetric flow rate: \[ v = v_0 \cdot [1 + 0.70 \cdot 0.40] \cdot \left(\frac{400}{300}\right) \] \[ v = v_0 \cdot [1 + 0.28] \cdot \left(\frac{4}{3}\right) = v_0 \cdot (1.28) \cdot \left(\frac{4}{3}\right) \] Convert the decimal value 1.28 to a fraction ($128/100 = 32/25$): \[ v = v_0 \cdot \left(\frac{32}{25}\right) \cdot \left(\frac{4}{3}\right) = v_0 \cdot \frac{128}{75} \]

Step 3: Calculating the ratio of outlet to inlet concentrations.
Let us write the expressions for the molar concentration of component $A$ at the inlet and outlet:

• Inlet Concentration: \[ C_{A0} = \frac{F_{A0}}{v_0} \]

• Outlet Concentration: \[ C_A = \frac{F_A}{v} = \frac{F_{A0} \cdot (1 - X_A)}{v} \]
Now, let us form the ratio of the outlet concentration to the inlet concentration: \[ \frac{C_A}{C_{A0}} = \frac{\left[\frac{F_{A0} \cdot (1 - X_A)}{v}\right]}{\left[\frac{F_{A0}}{v_0}\right]} = (1 - X_A) \cdot \left(\frac{v_0}{v}\right) \] Substitute the target conversion $X_A = 0.40$ and our previously derived volumetric flow rate ratio into this equation: \[ \frac{C_A}{C_{A0}} = (1 - 0.40) \cdot \left( \frac{1}{\frac{128}{75}} \right) = 0.60 \cdot \left( \frac{75}{128} \right) \] Convert the decimal value 0.60 to a fraction ($6/10 = 3/5$): \[ \frac{C_A}{C_{A0}} = \left(\frac{3}{5}\right) \cdot \left(\frac{75}{128}\right) = 3 \cdot \left(\frac{15}{128}\right) = \frac{45}{128} \] Let us calculate the final decimal value by performing long division: \[ \frac{45}{128} \approx 0.35156 \] Rounding this result to two decimal places yields exactly 0.35.