The first order reaction has a specific rate `10^(-3) s^(-1)` How much time will it take for 10 g of the reactant to reduce to 5 g ?

Options

A. 360 s
B. 180 s
C. 90 s
D. 270 s

Explanation

To solve the problem of how much time it will take for a first-order reaction to reduce 10 g of the reactant to 5 g, we can follow these steps:

### Step 1: Understand the first-order reaction kinetics
For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. The integrated rate law for a first-order reaction is given by:

\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]

where:
- \([A]_0\) is the initial concentration (or amount) of the reactant,
- \([A]\) is the concentration (or amount) of the reactant at time \(t\),
- \(k\) is the rate constant,
- \(t\) is the time.

### Step 2: Identify the values
From the problem:
- Initial amount, \([A]_0 = 10 \, \text{g}\)
- Final amount, \([A] = 5 \, \text{g}\)
- Rate constant, \(k = 10^{-3} \, \text{s}^{-1}\)

### Step 3: Substitute the values into the integrated rate law
Substituting the known values into the equation:

\[ \ln \left( \frac{10 \, \text{g}}{5 \, \text{g}} \right) = (10^{-3} \, \text{s}^{-1}) \cdot t \]

### Step 4: Simplify the equation
Calculate the left side:

\[ \ln(2) = (10^{-3}) \cdot t \]

### Step 5: Solve for \(t\)
Now, we need to find \(\ln(2)\):

\[ \ln(2) \approx 0.693 \]

Now substitute this value back into the equation:

\[ 0.693 = (10^{-3}) \cdot t \]

To find \(t\):

\[ t = \frac{0.693}{10^{-3}} = 693 \, \text{s} \]

### Conclusion
Thus, it will take approximately **693 seconds** for the reactant to reduce from 10 g to 5 g.

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