A first order reaction has a specific reaction rate of `10^(-2) s^(-1)`. How much time will it take for 20 g of the reactant to reduce to 5 g ?

Options

A. 2.38.6 second
B. 138.6 second
C. 346.5 second
D. 693.0 second

Answer

To solve the problem of how much time it will take for a first-order reaction to reduce 20 g of the reactant to 5 g, we can use the first-order rate equation. Here’s a step-by-step solution:

### Step 1: Understand the first-order reaction formula
For a first-order reaction, the relationship between the concentration of reactants and time is given by the equation:

\[ t = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A]} \right) \]

where:
- \( t \) is the time,
- \( k \) is the rate constant,
- \( [A_0] \) is the initial concentration (or mass in this case),
- \( [A] \) is the final concentration (or mass).

### Step 2: Identify the values
From the problem:
- The initial mass \( [A_0] \) = 20 g
- The final mass \( [A] \) = 5 g
- The rate constant \( k \) = \( 10^{-2} \, s^{-1} \)

### Step 3: Substitute the values into the equation
Now, substituting the values into the first-order rate equation:

\[ t = \frac{2.303}{10^{-2}} \log \left( \frac{20}{5} \right) \]

### Step 4: Calculate the logarithm
Calculate the logarithm:

\[ \frac{20}{5} = 4 \]
Thus, we need to find \( \log(4) \). We can express \( \log(4) \) as:

\[ \log(4) = 2 \log(2) \]

Using the known value \( \log(2) \approx 0.301 \):

\[ \log(4) = 2 \times 0.301 = 0.602 \]

### Step 5: Substitute back into the equation
Now substitute \( \log(4) \) back into the time equation:

\[ t = \frac{2.303}{10^{-2}} \times 0.602 \]

### Step 6: Calculate the time
Calculating the expression:

\[ t = 2.303 \times 100 \times 0.602 \]

\[ t = 230.3 \times 0.602 \approx 138.6 \, seconds \]

### Final Answer
Thus, the time it takes for the reaction to reduce 20 g of the reactant to 5 g is approximately **138.6 seconds**.

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