Select the rate law that corresponds to the datashown for the reaction `A+BtoC`
`{:(,Exp.,[A],[B],Rate),(,1,0.012,0.035,0.10),(,2,0.024,0.070,0.80),(,3,0.024,0.035,0.10),(,4,0.012,0.070,0.80):}`

Options

A. Rate=`k[B]^(3)`
B. Rate=`k[B]^(4)`
C. Rate k=`[A][B]^(3)`
D. Rate =k`[A]^(2)[B]^(2)`

Answer

To determine the rate law for the reaction \( A + B \to C \) based on the provided experimental data, we will analyze the data step by step.

### Step 1: Write the general rate law expression
The general rate law for the reaction can be expressed as:
\[ \text{Rate} = k[A]^x[B]^y \]
where \( k \) is the rate constant, \( [A] \) and \( [B] \) are the concentrations of reactants A and B, and \( x \) and \( y \) are the orders of the reaction with respect to A and B, respectively.

### Step 2: Analyze the data
We have the following experimental data:

| Exp. | [A] | [B] | Rate |
|------|-------|-------|-------|
| 1 | 0.012 | 0.035 | 0.10 |
| 2 | 0.024 | 0.070 | 0.80 |
| 3 | 0.024 | 0.035 | 0.10 |
| 4 | 0.012 | 0.070 | 0.80 |

### Step 3: Compare experiments to find x
Let's compare experiments 1 and 2.

From the data:
- For Experiment 1:
\[ \text{Rate}_1 = k[0.012]^x[0.035]^y = 0.10 \]
- For Experiment 2:
\[ \text{Rate}_2 = k[0.024]^x[0.070]^y = 0.80 \]

Now, we can set up the ratio of the rates:
\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[0.024]^x[0.070]^y}{k[0.012]^x[0.035]^y} \]
This simplifies to:
\[ \frac{0.80}{0.10} = \frac{[0.024]^x[0.070]^y}{[0.012]^x[0.035]^y} \]
\[ 8 = \frac{(0.024)^x(0.070)^y}{(0.012)^x(0.035)^y} \]

### Step 4: Solve for x and y
Now, we can simplify the equation:
\[ 8 = \left(\frac{0.024}{0.012}\right)^x \left(\frac{0.070}{0.035}\right)^y \]
Calculating the ratios:
\[ \frac{0.024}{0.012} = 2 \quad \text{and} \quad \frac{0.070}{0.035} = 2 \]
Thus, we have:
\[ 8 = 2^x \cdot 2^y = 2^{x+y} \]
This implies:
\[ x + y = 3 \quad \text{(Equation 1)} \]

### Step 5: Compare experiments to find y
Next, let's compare experiments 3 and 4.

From the data:
- For Experiment 3:
\[ \text{Rate}_3 = k[0.024]^x[0.035]^y = 0.10 \]
- For Experiment 4:
\[ \text{Rate}_4 = k[0.012]^x[0.070]^y = 0.80 \]

Setting up the ratio:
\[ \frac{\text{Rate}_4}{\text{Rate}_3} = \frac{k[0.012]^x[0.070]^y}{k[0.024]^x[0.035]^y} \]
This simplifies to:
\[ \frac{0.80}{0.10} = \frac{[0.012]^x[0.070]^y}{[0.024]^x[0.035]^y} \]
\[ 8 = \frac{(0.012)^x(0.070)^y}{(0.024)^x(0.035)^y} \]

### Step 6: Solve for x and y
This gives us:
\[ 8 = \left(\frac{0.012}{0.024}\right)^x \left(\frac{0.070}{0.035}\right)^y \]
Calculating the ratios:
\[ \frac{0.012}{0.024} = \frac{1}{2} \quad \text{and} \quad \frac{0.070}{0.035} = 2 \]
Thus, we have:
\[ 8 = \left(\frac{1}{2}\right)^x \cdot 2^y \]
This implies:
\[ 8 = 2^{-x + y} \]
Taking logarithm base 2 gives us:
\[ 3 = -x + y \quad \text{(Equation 2)} \]

### Step 7: Solve the system of equations
Now we have two equations:
1. \( x + y = 3 \)
2. \( -x + y = 3 \)

Adding these two equations:
\[ (x + y) + (-x + y) = 3 + 3 \]
\[ 2y = 6 \implies y = 3 \]

Substituting \( y = 3 \) into Equation 1:
\[ x + 3 = 3 \implies x = 0 \]

### Conclusion
The rate law for the reaction is:
\[ \text{Rate} = k[A]^0[B]^3 \]
This simplifies to:
\[ \text{Rate} = k[B]^3 \]