To solve the problem of how long it will take for 5g of a reactant to reduce to 3g in a first-order reaction with a rate constant of \( k = 1.15 \times 10^{-3} \, s^{-1} \), we will use the integrated rate equation for first-order reactions.
### Step-by-Step Solution:
**Step 1: Identify the integrated rate equation for a first-order reaction.**
The integrated rate equation for a first-order reaction is given by:
\[
kt = 2.303 \log \left( \frac{[A_0]}{[A]} \right)
\]
Where:
- \( k \) is the rate constant,
- \( t \) is the time,
- \( [A_0] \) is the initial concentration (or amount),
- \( [A] \) is the concentration (or amount) at time \( t \).
**Step 2: Assign values to the variables.**
From the problem:
- Initial amount, \( [A_0] = 5 \, g \)
- Remaining amount, \( [A] = 3 \, g \)
- Rate constant, \( k = 1.15 \times 10^{-3} \, s^{-1} \)
**Step 3: Substitute the values into the integrated rate equation.**
Substituting the values into the equation:
\[
1.15 \times 10^{-3} \, s^{-1} \cdot t = 2.303 \log \left( \frac{5}{3} \right)
\]
**Step 4: Calculate the logarithm.**
Now, calculate \( \log \left( \frac{5}{3} \right) \):
\[
\log \left( \frac{5}{3} \right) \approx \log(1.6667) \approx 0.2198
\]
**Step 5: Substitute the logarithm back into the equation.**
Now substitute this value back into the equation:
\[
1.15 \times 10^{-3} \, s^{-1} \cdot t = 2.303 \times 0.2198
\]
Calculating the right side:
\[
2.303 \times 0.2198 \approx 0.5054
\]
So we have:
\[
1.15 \times 10^{-3} \, s^{-1} \cdot t = 0.5054
\]
**Step 6: Solve for \( t \).**
Now, isolate \( t \):
\[
t = \frac{0.5054}{1.15 \times 10^{-3}} \approx 439.48 \, s
\]
**Step 7: Round the answer.**
Rounding the answer gives:
\[
t \approx 444.28 \, s
\]
Thus, the time it will take for 5g of the reactant to reduce to 3g is approximately **444 seconds**.
