The kinetic date for the given reaction `A(g)+2B(g)overset(K)toC(g)` is provided in the following table for three experiments at 300 K
`{:("Ex. No.","[A/M]","[B/M]","Initial rate" (M sec^(-1))),(1.,0.01,0.01,6.930xx10^(-6)),(2.,0.02,0.01,1.386xx10^(-5)),(3.,0.2,0.02,1.386xx10^(-5)):} `
In another experiment starting swith initial concentration of `0.5` and 1 M respectively for A and B at 300 K, find the rate of reaction after 50 minutes from the starting of edperiment (inM`//`sec).

Options

A. `6.93xx10^(-4)`
B. `0.25xx10^(-7)`
C. `4.33xx10^(-5)`
D. `3.46xx10^(-4)`

Answer

To solve the problem step by step, we will analyze the given data and apply the necessary calculations to find the rate of reaction after 50 minutes.

### Step 1: Analyze the given data
We have the following data from the experiments:

| Ex. No. | [A/M] | [B/M] | Initial Rate (M sec^(-1)) |
|---------|-------|-------|---------------------------|
| 1 | 0.01 | 0.01 | 6.930 x 10^(-6) |
| 2 | 0.02 | 0.01 | 1.386 x 10^(-5) |
| 3 | 0.2 | 0.02 | 1.386 x 10^(-5) |

### Step 2: Determine the rate law
From the data, we can determine the order of the reaction with respect to A and B.

1. **Comparing Experiments 1 and 2**:
- When [A] doubles from 0.01 M to 0.02 M (keeping [B] constant), the rate also approximately doubles.
- This suggests that the reaction is first order with respect to A.

2. **Comparing Experiments 2 and 3**:
- When [B] doubles from 0.01 M to 0.02 M (keeping [A] constant), the rate remains the same.
- This suggests that the reaction is zero order with respect to B.

Thus, the rate law can be expressed as:
\[ \text{Rate} = k[A]^1[B]^0 = k[A] \]

### Step 3: Calculate the rate constant (k)
Using the data from Experiment 1:
\[ \text{Rate} = k[A] \]
\[ 6.930 \times 10^{-6} = k \times 0.01 \]
\[ k = \frac{6.930 \times 10^{-6}}{0.01} = 6.93 \times 10^{-4} \, \text{M}^{-1} \text{s}^{-1} \]

### Step 4: Set up the equation for concentration of A over time
Using the first-order reaction formula:
\[ [A] = [A_0] e^{-kt} \]
Where:
- \([A_0] = 0.5 \, \text{M}\)
- \(k = 6.93 \times 10^{-4} \, \text{M}^{-1} \text{s}^{-1}\)
- \(t = 50 \, \text{minutes} = 50 \times 60 = 3000 \, \text{seconds}\)

Substituting the values:
\[ [A] = 0.5 e^{-6.93 \times 10^{-4} \times 3000} \]

### Step 5: Calculate the exponent
Calculate the exponent:
\[ -kt = -6.93 \times 10^{-4} \times 3000 = -2.079 \]

### Step 6: Calculate the concentration of A
Now substituting back:
\[ [A] = 0.5 e^{-2.079} \]
Calculating \(e^{-2.079}\):
\[ e^{-2.079} \approx 0.125 \]
Thus,
\[ [A] \approx 0.5 \times 0.125 = 0.0625 \, \text{M} \]

### Step 7: Calculate the rate of reaction
Using the rate law:
\[ \text{Rate} = k[A] \]
Substituting the values:
\[ \text{Rate} = 6.93 \times 10^{-4} \times 0.0625 \]
Calculating this gives:
\[ \text{Rate} \approx 4.33 \times 10^{-5} \, \text{M sec}^{-1} \]

### Final Answer
The rate of reaction after 50 minutes is approximately:
\[ \text{Rate} \approx 4.33 \times 10^{-5} \, \text{M sec}^{-1} \]

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