For the non-equilibrium process, `A + B rarr` Product, the rate is first-order w.r.t. `A` and second order w.r.t. `B`. If `1.0 mol` each of `A` and `B` were inrofuced into `1.0 L` vessel and the initial rate was `1.0 xx 10^(-2) mol L^(-1) s^(-1)`, calculate the rate when half the reactants have been turned into Products.

Options

A. `1.2 times 10^-5` mol/lt-sec
B. `1.2 times 10^-4` mol/lt-sec
C. `1.2 times 10^-3` mol/lt-sec
D. `1.2 times 10^-2` mol/lt -sec

Answer

To solve the problem step by step, we will follow the information provided in the question and the video transcript.

### Step 1: Write the Rate Law Expression
The rate law for the reaction \( A + B \rightarrow \text{Products} \) is given as:
\[ \text{Rate} = k [A]^1 [B]^2 \]
where \( k \) is the rate constant.

### Step 2: Determine Initial Concentrations
Since 1 mole of \( A \) and 1 mole of \( B \) are introduced into a 1.0 L vessel, the initial concentrations are:
\[ [A]_0 = 1.0 \, \text{mol/L} \]
\[ [B]_0 = 1.0 \, \text{mol/L} \]

### Step 3: Use the Initial Rate to Find the Rate Constant \( k \)
The initial rate is given as \( 1.0 \times 10^{-2} \, \text{mol L}^{-1} \text{s}^{-1} \). Plugging the initial concentrations into the rate law:
\[ 1.0 \times 10^{-2} = k (1.0)^1 (1.0)^2 \]
This simplifies to:
\[ 1.0 \times 10^{-2} = k \]
Thus, the rate constant \( k = 1.0 \times 10^{-2} \, \text{mol}^{-2} \text{L}^2 \text{s}^{-1} \).

### Step 4: Calculate Concentrations When Half the Reactants Are Used
When half of the reactants are converted into products, the remaining concentrations of \( A \) and \( B \) will be:
\[ [A] = \frac{1.0}{2} = 0.5 \, \text{mol/L} \]
\[ [B] = \frac{1.0}{2} = 0.5 \, \text{mol/L} \]

### Step 5: Calculate the New Rate
Now, we can calculate the new rate \( R' \) using the new concentrations:
\[ R' = k [A]^1 [B]^2 \]
Substituting the values:
\[ R' = (1.0 \times 10^{-2}) (0.5)^1 (0.5)^2 \]
Calculating \( (0.5)^2 = 0.25 \):
\[ R' = (1.0 \times 10^{-2}) (0.5) (0.25) \]
\[ R' = (1.0 \times 10^{-2}) (0.125) \]
\[ R' = 1.25 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Final Answer
The rate when half the reactants have been turned into products is:
\[ \boxed{1.25 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1}} \]