In a zero order reaction , 20% of the reaction complete in 10 s. How much time it will take to complete 50% of the reaction ?

Options

A. 20 s
B. 25 s
C. 30 s
D. 40 s

Answer

To solve the problem of how much time it will take to complete 50% of a zero-order reaction when 20% of the reaction is complete in 10 seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Zero-Order Reaction**:
In a zero-order reaction, the rate of reaction is constant and does not depend on the concentration of the reactants. The integrated rate law for a zero-order reaction is given by:
\[ [A] = [A_0] - kt \]
where \([A]\) is the concentration at time \(t\), \([A_0]\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time.

2. **Calculate the Rate Constant \(k\)**:
We know that 20% of the reaction is complete in 10 seconds. This means that 80% of the initial concentration remains. Therefore, we can express this as:
\[ [A] = 0.8[A_0] \]
Plugging this into the integrated rate law:
\[ 0.8[A_0] = [A_0] - k(10) \]
Rearranging gives:
\[ k(10) = [A_0] - 0.8[A_0] = 0.2[A_0] \]
Thus, we can solve for \(k\):
\[ k = \frac{0.2[A_0]}{10} = 0.02[A_0] \text{ (in concentration units per second)} \]

3. **Calculate Time for 50% Completion**:
For 50% completion, we have:
\[ [A] = 0.5[A_0] \]
Using the integrated rate law again:
\[ 0.5[A_0] = [A_0] - k(t_{50\%}) \]
Rearranging gives:
\[ k(t_{50\%}) = [A_0] - 0.5[A_0] = 0.5[A_0] \]
Substituting \(k\) into the equation:
\[ 0.02[A_0](t_{50\%}) = 0.5[A_0] \]
Dividing both sides by \([A_0]\) (assuming \([A_0] \neq 0\)):
\[ 0.02(t_{50\%}) = 0.5 \]
Solving for \(t_{50\%}\):
\[ t_{50\%} = \frac{0.5}{0.02} = 25 \text{ seconds} \]

### Final Answer:
The time it will take to complete 50% of the reaction is **25 seconds**.