The rate constant for a reaction is `("In 2")/(10) " min"^(-1)`. What will be the order of reaction and time taken to change concentration from 1 M to 0.25 M.

Options

A. One, 10 min
B. Zero, 10 min
C. One, 20 min
D. Two, 20 min

Explanation

To solve the problem step by step, we will determine the order of the reaction based on the given rate constant and then calculate the time taken for the concentration to change from 1 M to 0.25 M.

### Step 1: Identify the Order of the Reaction
The rate constant \( k \) is given as \( \frac{\ln 2}{10} \, \text{min}^{-1} \).

- The unit of the rate constant \( k \) is \( \text{min}^{-1} \).
- For a reaction, the units of the rate constant can help us determine the order:
- Zero order: \( \text{M min}^{-1} \)
- First order: \( \text{min}^{-1} \)
- Second order: \( \text{M}^{-1} \text{min}^{-1} \)

Since the unit of \( k \) is \( \text{min}^{-1} \), it indicates that the reaction is a **first-order reaction**.

### Step 2: Calculate the Time Taken for Concentration Change
We need to find the time taken for the concentration to change from 1 M to 0.25 M.

For a first-order reaction, the integrated rate law is given by:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]
Where:
- \([A]_0\) = initial concentration = 1 M
- \([A]\) = final concentration = 0.25 M
- \(k\) = rate constant = \(\frac{\ln 2}{10} \, \text{min}^{-1}\)
- \(t\) = time taken

Substituting the values into the equation:
\[ \ln \left( \frac{1}{0.25} \right) = \left( \frac{\ln 2}{10} \right) t \]

Calculating the left side:
\[ \ln \left( \frac{1}{0.25} \right) = \ln(4) = 2\ln(2) \]

Now substituting back into the equation:
\[ 2\ln(2) = \left( \frac{\ln 2}{10} \right) t \]

To isolate \( t \), we can divide both sides by \( \frac{\ln 2}{10} \):
\[ t = \frac{2\ln(2)}{\frac{\ln 2}{10}} = 2 \times 10 = 20 \, \text{minutes} \]

### Final Answer
- The order of the reaction is **first order**.
- The time taken to change the concentration from 1 M to 0.25 M is **20 minutes**.

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