Half-life period of a first order reaction is 10 min. Starting with initial concentration 12 M, the rate after 20 min is

Options

A. `0.693 "M min"^(-1)`
B. `0.693xx3 "M min"^(-1)`
C. `0.0693xx3 "M min"^(-1)`
D. `0.0693xx4 "M min"^(-1)`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the half-life of a first-order reaction
The half-life (t₁/₂) of a first-order reaction is given as 10 minutes. This means that after 10 minutes, the concentration of the reactant will be half of its initial concentration.

### Step 2: Determine the initial concentration
The initial concentration (C₀) is given as 12 M.

### Step 3: Calculate the concentration after the first half-life (10 minutes)
After the first half-life (10 minutes):
\[ C_1 = \frac{C_0}{2} = \frac{12 \, \text{M}}{2} = 6 \, \text{M} \]

### Step 4: Calculate the concentration after the second half-life (20 minutes)
After the second half-life (another 10 minutes, totaling 20 minutes):
\[ C_2 = \frac{C_1}{2} = \frac{6 \, \text{M}}{2} = 3 \, \text{M} \]

### Step 5: Calculate the rate constant (k)
The half-life of a first-order reaction is related to the rate constant (k) by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Rearranging this gives:
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{10 \, \text{min}} = 0.0693 \, \text{min}^{-1} \]

### Step 6: Calculate the rate of reaction after 20 minutes
The rate of a first-order reaction is given by:
\[ \text{Rate} = k \times [C] \]
Substituting the values we have:
\[ \text{Rate} = 0.0693 \, \text{min}^{-1} \times 3 \, \text{M} = 0.2079 \, \text{M/min} \]

### Final Answer
The rate of reaction after 20 minutes is approximately **0.2079 M/min**.

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