For the reaction, `2N_(2)O_(5)to4NO_(2)+O_(2)` rate and rate constant are `1.02xx10^(-4) M sec^(-1)` and `3.4xx10^(-5)sec^(-1)` respectively, the concentration of `N_(2)O_(5)`, at that time will be

Options

A. `3.4 xx 10^(-4)`
B. 3
C. 5.2
D. `3.2 xx 10^(-5)`

Explanation

To solve the problem, we need to find the concentration of \( N_2O_5 \) at a given time using the provided rate of the reaction and the rate constant. Here’s a step-by-step solution:

### Step 1: Identify the given values
We have the following information from the problem:
- Rate of the reaction \( = 1.02 \times 10^{-4} \, \text{M/s} \)
- Rate constant \( k = 3.4 \times 10^{-5} \, \text{s}^{-1} \)

### Step 2: Determine the order of the reaction
The units of the rate constant \( k \) are \( \text{s}^{-1} \), which indicates that the reaction is first-order with respect to \( N_2O_5 \).

### Step 3: Write the rate law expression
For a first-order reaction, the rate law can be expressed as:
\[ \text{Rate} = k \cdot [N_2O_5] \]
Where \( [N_2O_5] \) is the concentration of \( N_2O_5 \).

### Step 4: Rearrange the rate law to find the concentration of \( N_2O_5 \)
We can rearrange the rate law to solve for the concentration of \( N_2O_5 \):
\[ [N_2O_5] = \frac{\text{Rate}}{k} \]

### Step 5: Substitute the values into the equation
Now, we substitute the values of the rate and the rate constant into the equation:
\[ [N_2O_5] = \frac{1.02 \times 10^{-4} \, \text{M/s}}{3.4 \times 10^{-5} \, \text{s}^{-1}} \]

### Step 6: Calculate the concentration
Now, we perform the calculation:
\[ [N_2O_5] = \frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}} \approx 3.0 \, \text{M} \]

### Conclusion
The concentration of \( N_2O_5 \) at that time is approximately \( 3.0 \, \text{M} \).

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