The rate constant for an isomerization reaction, `A rarr B` is `4.5 xx 10^(-3) min^(-1)`. If the initial concentration of `A` is `1 M`, calculate the rate of the reaction after `1 h`.

Explanation

To solve the problem step by step, we will follow the principles of first-order kinetics.

### Step 1: Identify the given data
- Rate constant (k) = \(4.5 \times 10^{-3} \, \text{min}^{-1}\)
- Initial concentration of A ([A]₀) = 1 M
- Time (t) = 1 hour = 60 minutes

### Step 2: Use the first-order rate equation
For a first-order reaction, the relationship between the rate constant, time, and concentrations is given by:

\[ k = \frac{2.303}{t} \log\left(\frac{[A]_0}{[A]}\right) \]

Where:
- [A]₀ = initial concentration of A
- [A] = concentration of A at time t

### Step 3: Rearranging the equation
We can rearrange the equation to find the remaining concentration of A after time t:

\[ \log\left(\frac{[A]_0}{[A]}\right) = \frac{k \cdot t}{2.303} \]

### Step 4: Substitute the known values
Substituting the known values into the equation:

\[ \log\left(\frac{1}{[A]}\right) = \frac{(4.5 \times 10^{-3}) \cdot 60}{2.303} \]

### Step 5: Calculate the right-hand side
Calculating the right-hand side:

\[ \log\left(\frac{1}{[A]}\right) = \frac{0.27}{2.303} \approx 0.117 \]

### Step 6: Solve for [A]
Now, we can find [A]:

\[ \frac{1}{[A]} = 10^{0.117} \]

Calculating \(10^{0.117}\):

\[ \frac{1}{[A]} \approx 1.3 \]

Thus,

\[ [A] \approx \frac{1}{1.3} \approx 0.769 \, \text{M} \]

### Step 7: Calculate the rate of the reaction
The rate of the reaction can be calculated using the formula:

\[ \text{Rate} = k \times [A] \]

Substituting the values:

\[ \text{Rate} = (4.5 \times 10^{-3}) \times (0.769) \]

Calculating the rate:

\[ \text{Rate} \approx 3.45 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \]

### Final Answer
The rate of the reaction after 1 hour is approximately \(3.45 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1}\).

---