Question

A substrate is consumed in a zero order reaction such that its concentration falls from 42 g L\(^{-1}\) to 14 g L\(^{-1}\) in 4 hours. The total time taken for complete utilization of substrate will be hours. (answer in integer)

Hint:

Find the constant zero order rate from the given drop over 4 hours, then divide the full initial concentration by this rate.

Answer 1

Correct Answer: 6

Step 1: Understanding the Question.
In a zero order reaction, the rate of substrate consumption does not depend on how much substrate is left. So the concentration drops at a constant rate with time, giving a straight-line plot of concentration versus time.

Step 2: Key Formula.
For zero order kinetics, the concentration \(C\) at time \(t\) is
\[ C = C_0 - k_0 t \]
where \(C_0\) is the initial concentration and \(k_0\) is the (constant) zero order rate constant.

Step 3: Find the rate constant \(k_0\) from the given data.
The concentration falls from 42 g L\(^{-1}\) to 14 g L\(^{-1}\) in 4 hours, a drop of
\[ 42 - 14 = 28 \ \text{g L}^{-1} \]
so
\[ k_0 = \frac{28}{4} = 7 \ \text{g L}^{-1}\,\text{h}^{-1} \]

Step 4: Find the total time for complete utilization.
Complete utilization means the concentration falls all the way from \(C_0 = 42\) g L\(^{-1}\) to \(0\). Using the same constant rate,
\[ t_{total} = \frac{C_0}{k_0} = \frac{42}{7} = 6 \ \text{hours} \]

Final Answer:
\[ \boxed{t_{total} = 6 \ \text{hours}} \]