Rate expression for `Xa + Yb to` products is Rate `=K[A]^(m)[B]^(n)`. Units of K w.r.t A and B respectively are `s^(-1)` and `M^(-1)s^(-1)`, when concentrations of A and B are increased by 4 times, then

Options

A. `R_(f) = 16 R_(1)`
B. `R_(1) = 16R_(f)`
C. `R_(f) = 8R_(1)`
D. `R_(f) = 64 R_(i)`

Explanation

To solve the problem, we need to analyze the rate expression given and how the changes in concentrations affect the rate of the reaction.

### Step 1: Understand the Rate Expression
The rate expression for the reaction \( X_a + Y_b \rightarrow \text{products} \) is given as:
\[ \text{Rate} = k[A]^m[B]^n \]
where \( k \) is the rate constant, \( [A] \) and \( [B] \) are the concentrations of reactants \( A \) and \( B \), and \( m \) and \( n \) are the orders of the reaction with respect to \( A \) and \( B \), respectively.

### Step 2: Determine the Orders of Reaction
From the information provided:
- The units of \( k \) with respect to \( A \) are \( s^{-1} \).
- The units of \( k \) with respect to \( B \) are \( M^{-1}s^{-1} \).

Using the units of \( k \):
1. For \( A \):
\[ k \text{ (units)} = \text{M}^{0}(s^{-1}) \Rightarrow 1 - n = 0 \Rightarrow n = 1 \]
This indicates that the order with respect to \( A \) is \( m = 1 \).

2. For \( B \):
\[ k \text{ (units)} = \text{M}^{-1}(s^{-1}) \Rightarrow 1 - n = -1 \Rightarrow n = 2 \]
This indicates that the order with respect to \( B \) is \( n = 2 \).

### Step 3: Calculate the Initial Rate
Let’s assume the initial concentrations of \( A \) and \( B \) are \( [A] \) and \( [B] \). The initial rate \( R_i \) can be expressed as:
\[ R_i = k[A]^1[B]^2 = k[A][B]^2 \]

### Step 4: Change in Concentrations
If the concentrations of \( A \) and \( B \) are increased by 4 times, then:
\[ [A] \rightarrow 4[A] \]
\[ [B] \rightarrow 4[B] \]

### Step 5: Calculate the Final Rate
Now, substituting the new concentrations into the rate expression:
\[ R_f = k[4A][4B]^2 \]
Calculating this gives:
\[ R_f = k[4A][16B^2] = 64k[A][B]^2 \]

### Step 6: Relate Final Rate to Initial Rate
Since \( R_i = k[A][B]^2 \), we can express \( R_f \) in terms of \( R_i \):
\[ R_f = 64R_i \]

### Conclusion
Thus, when the concentrations of \( A \) and \( B \) are increased by 4 times, the final rate \( R_f \) becomes 64 times the initial rate \( R_i \).