If compound A reacts with B following first order kinetics with rate constant `2.011times10^(-3)s^(-1)`.The time taken by `A` (in seconds) to reduce from `7g` to `2g` will be..... `[log5=0.698,log7=0.845,log2=0.301]`

Answer

To solve the problem of how long it takes for compound A to reduce from 7 grams to 2 grams under first-order kinetics, we can use the first-order reaction formula:

### Step-by-Step Solution:

1. **Identify the Formula for First-Order Kinetics:**
The formula for the time (t) taken for a first-order reaction is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]} \right) \]
where:
- \( k \) is the rate constant,
- \( [A]_0 \) is the initial concentration (or amount) of A,
- \( [A] \) is the final concentration (or amount) of A.

2. **Substitute the Given Values:**
From the problem, we have:
- \( k = 2.011 \times 10^{-3} \, s^{-1} \)
- \( [A]_0 = 7 \, g \)
- \( [A] = 2 \, g \)

Plugging these values into the formula:
\[ t = \frac{2.303}{2.011 \times 10^{-3}} \log \left( \frac{7}{2} \right) \]

3. **Calculate the Logarithm:**
We can express the logarithm as:
\[ \log \left( \frac{7}{2} \right) = \log(7) - \log(2) \]
Using the provided logarithm values:
- \( \log(7) = 0.845 \)
- \( \log(2) = 0.301 \)

Therefore:
\[ \log \left( \frac{7}{2} \right) = 0.845 - 0.301 = 0.544 \]

4. **Substitute Back into the Time Formula:**
Now we substitute the logarithm value back into the time formula:
\[ t = \frac{2.303}{2.011 \times 10^{-3}} \times 0.544 \]

5. **Calculate the Time:**
First, calculate \( \frac{2.303}{2.011 \times 10^{-3}} \):
\[ \frac{2.303}{2.011 \times 10^{-3}} \approx 1145.2 \]
Now multiply by \( 0.544 \):
\[ t \approx 1145.2 \times 0.544 \approx 622.0 \, seconds \]

### Final Answer:
The time taken by A to reduce from 7 grams to 2 grams is approximately **622 seconds**.