In a zero order reaction half life is 100 sec. After how much time `(7)/(8)` fraction of reactant will be reacted?

Options

A. 300 sec.
B. 200 sec.
C. 175 sec.
D. 25 sec.

Answer

To solve the problem step by step, we will use the properties of zero-order reactions and the provided half-life information.

### Step 1: Understand the 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 relationship between the concentration of the reactant and time can be expressed as:
\[ [A] = [A_0] - kt \]
where:
- \([A]\) is the concentration of the reactant at time \(t\),
- \([A_0]\) is the initial concentration,
- \(k\) is the rate constant,
- \(t\) is the time.

### Step 2: Use the Half-Life Formula
The half-life (\(t_{1/2}\)) for a zero-order reaction is given by:
\[ t_{1/2} = \frac{[A_0]}{2k} \]
Given that the half-life is 100 seconds, we can set up the equation:
\[ 100 = \frac{[A_0]}{2k} \]

### Step 3: Solve for the Rate Constant \(k\)
Rearranging the half-life equation to find \(k\):
\[ 2k = \frac{[A_0]}{100} \]
\[ k = \frac{[A_0]}{200} \]

### Step 4: Determine the Amount of Reactant Reacted
We need to find the time required for \(\frac{7}{8}\) of the reactant to be reacted. If \(\frac{7}{8}\) of the reactant is reacted, then the remaining concentration \([A]\) will be:
\[ [A] = [A_0] - \frac{7}{8}[A_0] = \frac{1}{8}[A_0] \]

### Step 5: Substitute into the Zero-Order Equation
Now, substituting \([A]\) into the zero-order equation:
\[ \frac{1}{8}[A_0] = [A_0] - kt \]
Rearranging gives:
\[ kt = [A_0] - \frac{1}{8}[A_0] \]
\[ kt = \frac{7}{8}[A_0] \]

### Step 6: Substitute for \(k\)
Now substituting \(k = \frac{[A_0]}{200}\) into the equation:
\[ \frac{[A_0]}{200}t = \frac{7}{8}[A_0] \]

### Step 7: Cancel \([A_0]\) and Solve for \(t\)
Cancelling \([A_0]\) from both sides (assuming \([A_0] \neq 0\)):
\[ \frac{t}{200} = \frac{7}{8} \]
Now, solving for \(t\):
\[ t = 200 \times \frac{7}{8} \]
\[ t = 200 \times 0.875 \]
\[ t = 175 \text{ seconds} \]

### Conclusion
The time required for \(\frac{7}{8}\) of the reactant to be reacted is **175 seconds**.

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