146. A substrate is consumed in a zero order reaction such that the concentration falls from 40 g/l to 20 g/l in 4
h. How long will it take the substrate to fall from 20 g/l to 2 g/l:
(1) 2.5 h
(2) 1.8 h
(3) 3.6 h
(4) 4.8 h

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Detailed Explanation:

Question:
A substrate is consumed in a zero-order reaction such that the concentration falls from 40 g/l to 20 g/l in 4 hours. How long will it take the substrate to fall from 20 g/l to 2 g/l?

• (1) 2.5 h

• (2) 1.8 h

• (3) 3.6 h

• (4) 4.8 h

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Correct Answer:

(3) 3.6 h

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Explanation:

In a zero-order reaction, the rate of consumption of the substrate is constant, which means the rate of change in concentration of the substrate is independent of its current concentration. This is described by the rate equation:

d[A]dt=−k\frac{d[A]}{dt} = -kdtd[A]=−k

Where:

• $[A]$ is the concentration of the substrate.

• $k$ is the rate constant (units of concentration/time).

The general integrated form of the zero-order rate law is:

[A]=[A0]−kt[A] = [A_0] – kt[A]=[A0]−kt

Where:

• [A0][A_0][A0] is the initial concentration.

• $[A]$ is the concentration at time $t$.

• $k$ is the rate constant.

• $t$ is time.

Step 1: Determine the rate constant, $k$:

We know the concentration falls from 40 g/l to 20 g/l in 4 hours. Using the integrated rate equation:

[A]=[A0]−kt[A] = [A_0] – kt[A]=[A0]−kt

Substituting the given values:

$$20=40-k\times 4$$

Solving for $k$:

$$20=40-4k\Rightarrow 4k=20\Rightarrow k=5\text{ g/l/h}$$

Step 2: Use the rate constant to calculate the time for the concentration to fall from 20 g/l to 2 g/l:

Now, using the same equation for the next time interval, where the initial concentration is 20 g/l and the final concentration is 2 g/l:

[A]=[A0]−kt[A] = [A_0] – kt[A]=[A0]−kt

Substituting the known values:

$$2=20-5\times t$$

Solving for $t$:

$$2=20-5t\Rightarrow 5t=18\Rightarrow t=3.6\text{ hours}$$

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Conclusion:

The time it will take for the substrate to fall from 20 g/l to 2 g/l in this zero-order reaction is 3.6 hours, as calculated using the rate constant derived from the initial data. Therefore, the correct answer is (3) 3.6 h.