For the reaction `2N_2O_5 to 4NO_2 +O_2` rate of reaction and rate constant are `1.04 xx 10^(-4)and 3.4 xx 10^(-5)sec^(-1)` respectively. The concentration of `N_2O_5` at that time will be :

Options

A. 1.732
B. 3
C. 3.06
D. `3.4 xx 10^(-5)`

Answer

To find the concentration of \( N_2O_5 \) at the given time for the reaction \( 2N_2O_5 \rightarrow 4NO_2 + O_2 \), we can follow these steps:

### Step 1: Identify the order of the reaction
Given that the rate constant \( k \) has units of \( \text{s}^{-1} \), we can conclude that this is a first-order reaction. In a first-order reaction, the rate is directly proportional to the concentration of the reactant.

### Step 2: Write the rate law expression
For a first-order reaction, the rate law can be expressed as:
\[ \text{Rate} = k [N_2O_5] \]
where:
- Rate is the rate of the reaction,
- \( k \) is the rate constant,
- \( [N_2O_5] \) is the concentration of the reactant \( N_2O_5 \).

### Step 3: Substitute the known values
We know:
- Rate \( = 1.04 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \)
- Rate constant \( k = 3.4 \times 10^{-5} \, \text{s}^{-1} \)

Now, substituting these values into the rate law:
\[ 1.04 \times 10^{-4} = 3.4 \times 10^{-5} [N_2O_5] \]

### Step 4: Solve for the concentration of \( N_2O_5 \)
To find \( [N_2O_5] \), rearrange the equation:
\[ [N_2O_5] = \frac{1.04 \times 10^{-4}}{3.4 \times 10^{-5}} \]

### Step 5: Calculate the concentration
Now, performing the calculation:
\[ [N_2O_5] = \frac{1.04}{3.4} \times 10^{-4 + 5} = \frac{1.04}{3.4} \times 10^{1} \]
Calculating \( \frac{1.04}{3.4} \):
\[ \frac{1.04}{3.4} \approx 0.30588 \]
Thus,
\[ [N_2O_5] \approx 0.30588 \times 10^{1} \approx 3.0588 \, \text{mol L}^{-1} \]

### Final Answer
The concentration of \( N_2O_5 \) at that time will be approximately:
\[ [N_2O_5] \approx 3.06 \, \text{mol L}^{-1} \]

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