A first order reaction has a rate constant of `5 xx 10^(-3) s^(-1)`. How long will `5.0 g` of this reaction take to reduce to `3.0 g` ?

Options

A. 34.07 s
B. 7.57 s
C. 10.10 s
D. 15 g

Explanation

To solve the problem of how long it will take for a first-order reaction to reduce from 5.0 g to 3.0 g with a rate constant of \(5 \times 10^{-3} \, s^{-1}\), we can use the first-order kinetics equation. Here’s the step-by-step solution:

### Step 1: Identify the variables
- Initial amount (A) = 5.0 g
- Final amount (A - x) = 3.0 g
- Amount reacted (x) = A - (A - x) = 5.0 g - 3.0 g = 2.0 g
- Rate constant (k) = \(5 \times 10^{-3} \, s^{-1}\)

### Step 2: Use the first-order reaction formula
The first-order kinetics equation is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{A}{A - x} \right) \]
Substituting the known values into the equation:
\[ t = \frac{2.303}{5 \times 10^{-3}} \log \left( \frac{5.0}{3.0} \right) \]

### Step 3: Calculate the logarithm
First, we need to calculate \(\log \left( \frac{5.0}{3.0} \right)\):
\[ \log \left( \frac{5.0}{3.0} \right) = \log(5.0) - \log(3.0) \]
Using logarithm values:
- \(\log(5.0) \approx 0.6990\)
- \(\log(3.0) \approx 0.4771\)

Now, calculate:
\[ \log \left( \frac{5.0}{3.0} \right) = 0.6990 - 0.4771 = 0.2219 \]

### Step 4: Substitute back into the equation
Now substitute this value back into the equation for \(t\):
\[ t = \frac{2.303}{5 \times 10^{-3}} \times 0.2219 \]

### Step 5: Calculate \(t\)
First, calculate the denominator:
\[ 5 \times 10^{-3} = 0.005 \]
Now calculate:
\[ t = \frac{2.303 \times 0.2219}{0.005} \]
Calculating \(2.303 \times 0.2219 \approx 0.5113\):
\[ t = \frac{0.5113}{0.005} = 102.26 \, s \]

### Step 6: Final calculation
Now, we can finalize the calculation:
\[ t \approx 102.26 \, s \]

### Conclusion
The time taken for the reaction to reduce from 5.0 g to 3.0 g is approximately **102.26 seconds**.

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