Question

The catalytic gas-phase reaction \( A \rightarrow {products} \) is carried out in an isothermal batch reactor of 10 L volume using 0.1 kg of a solid catalyst. The reaction is first-order with: \[ (-r_A) = k' a(t) C_A \] Where: - \( k' = \frac{1}{{kg catalyst} \cdot {h}} \) - \( C_A \) is the concentration of \( A \) in mol/L. The catalyst activity \( a(t) \) undergoes first-order decay with rate constant \( k_d = 0.01 \) per hour and \( a(0) = 1 \). The reactant conversion after 1 day of operation is __________ (rounded off to 2 decimal places).

Hint:

For first-order reactions with decaying catalyst activity, use the activity decay model to calculate the conversion over time.

Answer 1

The catalyst activity \( a(t) \) decays according to the first-order rate law: \[ a(t) = a(0) e^{-k_d t} \] Given: - \( k_d = 0.01 \, {h}^{-1} \), - \( t = 24 \, {hours} \), - \( a(0) = 1 \). The activity at \( t = 24 \) hours is: \[ a(24) = e^{-0.01 \times 24} = e^{-0.24} \approx 0.7866 \] The concentration \( C_A(t) \) at time \( t \) is related to the initial concentration by the following equation: \[ C_A(t) = \frac{C_A(0)}{1 + k' a(t) t} \] The conversion \( X \) is given by: \[ X = 1 - \frac{C_A(t)}{C_A(0)} = 1 - \frac{1}{1 + k' a(t) t} \] Substitute the values: \[ X = 1 - \frac{1}{1 + 10 \times 0.7866 \times 24} \approx 0.17 \] Thus, the conversion after 1 day of operation is 0.17.