For the reaction `2A+BrarrC`, the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reaction is : `{:([A]("mol"L^(-1)),[B]("mol"L^(-1)),{:("Initial Rate"),("mol"^(-1)s^(-1)):}),(0.05,0.05,0.045),(0.10,0.05,0.090),(0.20,0.10,0.72):}`

Options

A. Rate`=K[A][B]^(2)`
B. Rate=`k[A]^(2)[B]^(2)`
C. Rate`=K[A][B]`
D. Rate `=k[A]^(2)[B]`

Answer

To determine the rate law for the reaction \(2A + B \rightarrow C\) based on the provided initial rates and concentrations, we will follow these steps:

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

### Step 2: Analyze the data from the table
The initial rates and concentrations are given as follows:

| [A] (mol L\(^{-1}\)) | [B] (mol L\(^{-1}\)) | Initial Rate (mol L\(^{-1}\) s\(^{-1}\)) |
|-----------------------|-----------------------|-------------------------------------------|
| 0.05 | 0.05 | 0.045 |
| 0.10 | 0.05 | 0.090 |
| 0.20 | 0.10 | 0.72 |

### Step 3: Set up equations using the data
Using the first experiment:
\[ 0.045 = k (0.05)^p (0.05)^q \quad \text{(Equation 1)} \]

Using the second experiment:
\[ 0.090 = k (0.10)^p (0.05)^q \quad \text{(Equation 2)} \]

Using the third experiment:
\[ 0.72 = k (0.20)^p (0.10)^q \quad \text{(Equation 3)} \]

### Step 4: Divide equations to eliminate \(k\)
To find \(p\), divide Equation 2 by Equation 1:
\[ \frac{0.090}{0.045} = \frac{k (0.10)^p (0.05)^q}{k (0.05)^p (0.05)^q} \]
This simplifies to:
\[ 2 = \frac{(0.10)^p}{(0.05)^p} \implies 2 = 2^p \implies p = 1 \]

### Step 5: Find \(q\)
Now, we will use the value of \(p\) to find \(q\). Divide Equation 3 by Equation 2:
\[ \frac{0.72}{0.090} = \frac{k (0.20)^p (0.10)^q}{k (0.10)^p (0.05)^q} \]
This simplifies to:
\[ 8 = \frac{(0.20)^1 (0.10)^q}{(0.10)^1 (0.05)^q} \]
\[ 8 = \frac{0.20}{0.10} \cdot \frac{(0.10)^q}{(0.05)^q} \implies 8 = 2 \cdot \left(\frac{0.10}{0.05}\right)^q \]
\[ 8 = 2 \cdot 2^q \implies 4 = 2^q \implies q = 2 \]

### Step 6: Write the final rate law
Now that we have \(p = 1\) and \(q = 2\), we can write the rate law:
\[ \text{Rate} = k [A]^1 [B]^2 \implies \text{Rate} = k [A] [B]^2 \]

### Conclusion
The rate law for the reaction \(2A + B \rightarrow C\) is:
\[ \text{Rate} = k [A] [B]^2 \]