For the reaction `A + 2B rarr` products (started with concentration taken in stoichiometric proportion), the experimentally determined rate law is :
`-(d[A])/(d t) = ksqrt([A])sqrt([B])`
The half life time of the reaction would be :

Options

A. `(0.693)/(k)`
B. `(0.693)/(1//k)`
C. `(0.693)/(sqrt(2k))`
D. not defined

Explanation

To find the half-life of the reaction \( A + 2B \rightarrow \text{products} \) with the given rate law:

\[ -\frac{d[A]}{dt} = k \sqrt{[A]} \sqrt{[B]} \]

we will follow these steps:

### Step 1: Understand the stoichiometric proportions
Since the reaction starts with concentrations in stoichiometric proportions, we can assume that initially, the concentration of \( A \) is \( [A]_0 \) and the concentration of \( B \) is \( 2[A]_0 \) (because it requires 2 moles of \( B \) for every mole of \( A \)).

### Step 2: Write the rate law in terms of concentration
From the rate law, we have:

\[ -\frac{d[A]}{dt} = k \sqrt{[A]} \sqrt{[B]} = k \sqrt{[A]} \sqrt{2[A]_0 - [A]} \]

### Step 3: Substitute \( [B] \) in terms of \( [A] \)
Since \( [B] = 2[A]_0 - [A] \), we can substitute this into the rate equation:

\[ -\frac{d[A]}{dt} = k \sqrt{[A]} \sqrt{2[A]_0 - [A]} \]

### Step 4: Rearrange the equation
Rearranging gives us:

\[ \frac{d[A]}{\sqrt{[A]} \sqrt{2[A]_0 - [A]}} = -k dt \]

### Step 5: Integrate the equation
Integrate both sides. The left side requires a suitable substitution or recognition of the integral form. The limits for \( [A] \) will go from \( [A]_0 \) to \( [A]_0/2 \) for half-life.

### Step 6: Find the expression for half-life
After integration, we find that the half-life \( t_{1/2} \) can be expressed as:

\[ t_{1/2} = \frac{0.693}{k \sqrt{2[A]_0}} \]

### Final Expression
Thus, the half-life of the reaction is:

\[ t_{1/2} = \frac{0.693}{k \sqrt{2[A]_0}} \]