`[A](M)" "[B](M)" ""Initial rate"(Ms^(-1))`
`0.4" "0.3" "2xx10^(-3)`
`0.8" "0.3" "0.8xx10^(-2)`
`1.2" "0.9" "0.54xx10^(-1)`
From the above data the rate law for the equation `A+Bto` products is equal to

Options

A. `"K[A][B]"`
B. `"K[A]"^(2)"[B]"^(2)`
C. `"K[A]"^(2)"[B]"`
D. `"K[A][B]"^(2)`

Answer

To determine the rate law for the reaction \( A + B \to \text{products} \) based on the provided data, we will follow these steps:

### Step 1: Write the Rate Law Expression
The general rate law can be expressed as:
\[ \text{Rate} = k [A]^x [B]^y \]
where \( k \) is the rate constant, \( x \) is the order with respect to \( A \), and \( y \) is the order with respect to \( B \).

### Step 2: Set Up Equations from the Given Data
We will use the initial rates and concentrations from the table to create equations.

1. From the first set of data:
\[ 2 \times 10^{-3} = k (0.4)^x (0.3)^y \quad \text{(Equation 1)} \]

2. From the second set of data:
\[ 0.8 \times 10^{-2} = k (0.8)^x (0.3)^y \quad \text{(Equation 2)} \]

3. From the third set of data:
\[ 0.54 \times 10^{-1} = k (1.2)^x (0.9)^y \quad \text{(Equation 3)} \]

### Step 3: Divide Equations to Eliminate \( k \)
To find \( x \) and \( y \), we can divide the equations to eliminate \( k \).

**Dividing Equation 1 by Equation 2:**
\[ \frac{2 \times 10^{-3}}{0.8 \times 10^{-2}} = \frac{k (0.4)^x (0.3)^y}{k (0.8)^x (0.3)^y} \]
This simplifies to:
\[ \frac{2 \times 10^{-3}}{0.8 \times 10^{-2}} = \frac{(0.4)^x}{(0.8)^x} \]
Calculating the left side:
\[ \frac{2 \times 10^{-3}}{0.8 \times 10^{-2}} = \frac{2}{0.8} \times 10^{-1} = 2.5 \times 10^{-1} \]
Thus, we have:
\[ 2.5 \times 10^{-1} = \left(\frac{0.4}{0.8}\right)^x = (0.5)^x \]
This implies:
\[ 2.5 = 5^x \quad \text{(since } 10^{-1} \text{ cancels out)} \]
Taking logarithm:
\[ x = 2 \]

**Dividing Equation 3 by Equation 2:**
\[ \frac{0.54 \times 10^{-1}}{0.8 \times 10^{-2}} = \frac{k (1.2)^x (0.9)^y}{k (0.8)^x (0.3)^y} \]
This simplifies to:
\[ \frac{0.54 \times 10^{-1}}{0.8 \times 10^{-2}} = \frac{(1.2)^x (0.9)^y}{(0.8)^x (0.3)^y} \]
Calculating the left side:
\[ \frac{0.54 \times 10^{-1}}{0.8 \times 10^{-2}} = \frac{0.54}{0.8} \times 10 = 6.75 \]
Thus, we have:
\[ 6.75 = \frac{(1.2)^x (0.9)^y}{(0.8)^x (0.3)^y} \]
Substituting \( x = 2 \):
\[ 6.75 = \frac{(1.2)^2 (0.9)^y}{(0.8)^2 (0.3)^y} \]
Calculating \( (1.2)^2 \) and \( (0.8)^2 \):
\[ 6.75 = \frac{1.44 (0.9)^y}{0.64 (0.3)^y} \]
Cross-multiplying gives:
\[ 6.75 \times 0.64 (0.3)^y = 1.44 (0.9)^y \]
Calculating \( 6.75 \times 0.64 \):
\[ 4.32 (0.3)^y = 1.44 (0.9)^y \]
Rearranging gives:
\[ \frac{(0.3)^y}{(0.9)^y} = \frac{1.44}{4.32} \]
This simplifies to:
\[ \left(\frac{0.3}{0.9}\right)^y = \frac{1}{3} \]
Thus:
\[ y = 1 \]

### Step 4: Write the Final Rate Law
Now that we have \( x = 2 \) and \( y = 1 \), the rate law can be expressed as:
\[ \text{Rate} = k [A]^2 [B]^1 \]

### Final Answer:
The rate law for the reaction \( A + B \to \text{products} \) is:
\[ \text{Rate} = k [A]^2 [B] \]

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