A first order reaction has a rate constant `1.15xx10^(-3)s^(-1)`. How long will `5g` of this reactant take to reduce to `3g`` ?

Options

A. 444s
B. 400s
C. 528s
D. 669s

Explanation

To solve the problem of how long it takes for a first-order reaction to reduce the amount of a reactant from 5 g to 3 g, we will use the first-order kinetics formula. The formula for the time \( t \) in a first-order reaction is given by:

\[ t = \frac{2.303}{k} \log \left( \frac{R_0}{R} \right) \]

Where:
- \( t \) is the time,
- \( k \) is the rate constant,
- \( R_0 \) is the initial concentration (or amount) of the reactant,
- \( R \) is the final concentration (or amount) of the reactant.

### Step 1: Identify the given values
- Rate constant \( k = 1.15 \times 10^{-3} \, s^{-1} \)
- Initial amount \( R_0 = 5 \, g \)
- Final amount \( R = 3 \, g \)

### Step 2: Substitute the values into the formula
Using the formula, we substitute the values we have:

\[ t = \frac{2.303}{1.15 \times 10^{-3}} \log \left( \frac{5}{3} \right) \]

### Step 3: Calculate the ratio \( \frac{R_0}{R} \)
Calculate the ratio of the initial amount to the final amount:

\[ \frac{R_0}{R} = \frac{5}{3} \approx 1.6667 \]

### Step 4: Calculate the logarithm
Now, we need to calculate the logarithm of the ratio:

\[ \log(1.6667) \approx 0.2218 \]

### Step 5: Substitute back into the time equation
Now, substitute the logarithm back into the time equation:

\[ t = \frac{2.303}{1.15 \times 10^{-3}} \times 0.2218 \]

### Step 6: Calculate the time \( t \)
Now, perform the calculation:

\[ t \approx \frac{2.303 \times 0.2218}{1.15 \times 10^{-3}} \approx \frac{0.5110}{1.15 \times 10^{-3}} \approx 444.35 \, seconds \]

### Step 7: Round the answer
Rounding to the nearest whole number, we get:

\[ t \approx 444 \, seconds \]

### Final Answer
The time it takes for 5 g of the reactant to reduce to 3 g is approximately **444 seconds**.

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