Half life period of a first order reaction is 10 minutes. Starting with 10 M, rate after 20 minutes is

Options

A. `0.0693 "M min"^(-1)`
B. `0.0693 xx2.5 "M min"^(-1)`
C. `0.0693 xx 5 "M min"^(-1)`
D. `0.0693xx10 M "min"^(-1)`

Explanation

To solve the problem, 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 every 10 minutes, the concentration of the reactant will reduce to half of its previous value.

### Step 2: Calculate the concentration after 20 minutes
Since the half-life is 10 minutes, we can determine how many half-lives fit into 20 minutes:
- After the first 10 minutes (1 half-life), the concentration will be:
\[ \text{Concentration after 10 minutes} = \frac{10 \, \text{M}}{2} = 5 \, \text{M} \]
- After the second 10 minutes (2 half-lives), the concentration will be:
\[ \text{Concentration after 20 minutes} = \frac{5 \, \text{M}}{2} = 2.5 \, \text{M} \]

### Step 3: Calculate the rate constant (k)
The rate constant (k) for a first-order reaction can be calculated using the formula:
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting the half-life:
\[ k = \frac{0.693}{10 \, \text{minutes}} = 0.0693 \, \text{min}^{-1} \]

### Step 4: Calculate the rate of the reaction
The rate of a first-order reaction is given by the equation:
\[ \text{Rate} = k \times [A] \]
Where [A] is the concentration of the reactant at the time of interest. Substituting the values we have:
\[ \text{Rate} = 0.0693 \, \text{min}^{-1} \times 2.5 \, \text{M} = 0.17325 \, \text{M min}^{-1} \]

### Final Answer
The rate after 20 minutes is approximately **0.173 M min⁻¹**.

---