GATE 2021 CH – Question 60
Reactant A decomposes to products B and C in the presence of an enzyme in a well-stirred batch reactor. The kinetic rate expression is $-r_A=\frac{0.01C_A}{0.05+C_A}$ (mol L⁻¹ min⁻¹). If the initial concentration of A is 0.02 mol/L, the time taken to achieve 50% conversion of A is _____ min (round off to 2 decimal places).
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Correct answer: 4.45 to 4.49
Explanation
At 50% conversion $C_{Af}=0.01$ mol/L. For a constant-volume batch reactor, $-dC_A/dt=0.01C_A/(0.05+C_A)$. Separate variables and integrate:
$$t=\frac1{0.01}\int_{0.01}^{0.02}\left(\frac{0.05}{C_A}+1\right)dC_A$$
$$=\frac{0.05\ln(0.02/0.01)+(0.02-0.01)}{0.01}=4.4657359\ \mathrm{min}.$$
Thus **t = 4.47 min**. Retaining both terms avoids incorrectly treating the enzyme kinetics as purely zero-order or first-order.