Question

The half-life of a substance in a certain enzyme-catalyzed reaction is $\text{ 138 s }$. The time required for the concentration of the substance to fall from $\text{ 1}\text{.28 }$to $\text{ 0}\text{.04 mg}{{\text{L}}^{-1}}\text{ }$ is:
A) $\text{ 414 s }$
B) $\text{ 522 s }$
C) $\text{ 690 s }$
D) $\text{ 276 s }$

Answer 1

The enzyme-catalyzed reaction follows the first-order reaction. The integrated rate law for the first-order reaction is as shown below,
$\text{ }{{\text{k}}_{\text{1}}}\text{ = }\dfrac{1}{t}\text{ ln }\dfrac{a}{a-x}\text{ = }\dfrac{2.303}{t}\text{ log }\dfrac{{{\left[ \text{A} \right]}_{0}}}{{{\left[ \text{A} \right]}_{t}}}\text{ }$

Where $\text{ }{{\text{k}}_{\text{1}}}\text{ }$ is the rate constant of the first-order reaction, ${{\left[ \text{A} \right]}_{0}}$ is initial concentration, and ${{\left[ \text{A} \right]}_{t}}$ is the concentration of a reactant at a time ‘t’. The half-life period for the first-order reaction is when $\text{ 50}{\scriptstyle{}^{0}/{}_{0}}\text{ }$ of the initial concentration of the reactant is converted into the product.

Complete step by step answer:
Enzymes are proteins with high molecular weight and are derived from living organisms. They are used as the homogenous catalyst in various organic reactions. The enzymes catalyzed reactions are likely to follow the first-order kinetic reaction.
The reaction is said to follow first-order kinetics if the rate of reaction depends on the concentration of one reactant only. The integrated rate expression for the first-order reaction is as shown below,
$\text{ }{{\text{k}}_{\text{1}}}\text{ = }\dfrac{1}{t}\text{ ln }\dfrac{a}{a-x}\text{ = }\dfrac{2.303}{t}\text{ log }\dfrac{{{\left[ \text{A} \right]}_{0}}}{{{\left[ \text{A} \right]}_{t}}}\text{ }$ (1)
Where $\text{ }{{\text{k}}_{\text{1}}}\text{ }$ is the rate constant of the first-order reaction, ${{\left[ \text{A} \right]}_{0}}$ is initial concentration, and ${{\left[ \text{A} \right]}_{t}}$ is the concentration of a reactant at a time ‘t’.

The half-life of the first-order reaction is equal to,
$\text{ }{{t}_{\dfrac{1}{2}}}\text{ = }\dfrac{0.693}{{{k}_{1}}}\text{ }$
Where, ${{t}_{1/2}}\text{ }$ is the half-life period for the first-order reaction.

We have given the following data:
The half lifetime period ${{t}_{1/2}}\text{ }$ is given as,$\text{ }{{t}_{1/2}}\text{ = 138 sec }$
The initial concentration of a substance, $\text{ }{{\left[ \text{A} \right]}_{0}}\text{ = 1}\text{.28 mg}{{\text{L}}^{-1}}\text{ }$
The final concentration of a substance, $\text{ }{{\left[ \text{A} \right]}_{\text{t}}}\text{ = 0}\text{.04 mg}{{\text{L}}^{-1}}\text{ }$
We are interested to find the time required for the change in concentration.

Let's first find out the rate constant $\text{ }{{\text{k}}_{\text{1}}}\text{ }$ from ${{t}_{1/2}}\text{ }$the equation. We have,

& \text{ }{{\text{t}}_{\text{1/2}}}\text{ = }\frac{0.693}{{{k}_{1}}}\text{ } \\\ & \Rightarrow {{k}_{1}}\text{ = }\dfrac{0.693}{{{\text{t}}_{\text{1/2}}}}\text{ = }\dfrac{0.693}{138}\text{ = 5}\text{.021 }\times \text{1}{{\text{0}}^{-3}}\text{ se}{{\text{c}}^{-1}}\text{ } \\\ \end{aligned}$$ Since we have calculated the value of the rate constant. Let's just substitute in the integrated rate law of first-order reaction (1) on rearrangement with respect to time ‘t’ we have, $\begin{aligned} & \text{ }t\text{ = }\dfrac{2.303}{{{\text{k}}_{\text{1}}}}\text{ log }\dfrac{{{\left[ \text{A} \right]}_{0}}}{{{\left[ \text{A} \right]}_{t}}} \\\ & \Rightarrow \dfrac{2.303}{\text{5}\text{.021 }\times \text{1}{{\text{0}}^{-3}}\text{ se}{{\text{c}}^{-1}}\text{ }}\text{ log }\dfrac{1.28}{0.04} \\\ & \Rightarrow \left( 458.60 \right)\times \left( \log \text{ 32} \right) \\\ & \Rightarrow \left( 458.60 \right)\left( 1.505 \right) \\\ & \therefore t\text{ = 690 sec } \\\ \end{aligned}$ Thus, the time required for the conversion of a substance from the $\text{ }{{\left[ \text{A} \right]}_{0}}\text{ = 1}\text{.28 mg}{{\text{L}}^{-1}}\text{ }$to $\text{ }{{\left[ \text{A} \right]}_{\text{t}}}\text{ = 0}\text{.04 mg}{{\text{L}}^{-1}}\text{ }$is equal to $\text{ 690 sec }$. **So, the correct answer is “Option C”.** **Note:** Note that, the kinetics of the enzyme-catalyzed reaction is proposed by Michaelis Menten equation. According to which the reaction rate of enzyme-catalyzed reaction changes from the first order to zero order as the substrate concentration is increased. This is because enzyme molecules have more number of active sites. Here, in this question, we have assumed the reaction follows first-order kinetics.