For reaction `2N_2 O_5 = 2NO_2 + O_2`, the rate and rate constants are `1.02 xx 10^(-4)` mole litre`""^(-1) sec^(-1) and 3.4 xx 10^(-5) sec^(-1)` respectively. The concentration of `N_2 O_5` at that time will be

Options

A. `1.732 mol lit^(-1)`
B. `3 mol L^(-1)`
C. `1.02 xx 10^(-4) mol Lit^(-1)`
D. `3.2 xx 10^5 mol lit^(-1)`

Explanation

To solve the problem, we need to find the concentration of \( N_2O_5 \) at a given time for the reaction:

\[ 2N_2O_5 \rightarrow 2NO_2 + O_2 \]

Given data:
- Rate of reaction = \( 1.02 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \)
- Rate constant \( k = 3.4 \times 10^{-5} \, \text{s}^{-1} \)

### Step 1: Determine the order of the reaction

The unit of the rate constant \( k \) is given as \( \text{s}^{-1} \). For a reaction of order \( n \), the unit of \( k \) can be expressed as:

\[ \text{Unit of } k = \text{mol L}^{-1} \text{s}^{-1} \times \text{(mol L}^{-1})^{1-n} \]

Since \( k \) has units of \( \text{s}^{-1} \), we can set up the equation:

\[ \text{s}^{-1} = \text{mol L}^{-1} \times \text{(mol L}^{-1})^{1-n} \]

This implies:

\[ 1 - n = 0 \quad \Rightarrow \quad n = 1 \]

Thus, the reaction is first order with respect to \( N_2O_5 \).

### Step 2: Write the rate equation

For a first-order reaction, the rate equation can be expressed as:

\[ \text{Rate} = k [N_2O_5] \]

### Step 3: Rearrange the equation to find the concentration of \( N_2O_5 \)

We can rearrange the rate equation to solve for the concentration of \( N_2O_5 \):

\[ [N_2O_5] = \frac{\text{Rate}}{k} \]

### Step 4: Substitute the values into the equation

Now, we can substitute the given values into the equation:

\[ [N_2O_5] = \frac{1.02 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1}}{3.4 \times 10^{-5} \, \text{s}^{-1}} \]

### Step 5: Calculate the concentration

Calculating the concentration:

\[ [N_2O_5] = \frac{1.02}{3.4} \times 10^{-4 + 5} = \frac{1.02}{3.4} \times 10^{1} \]

Calculating \( \frac{1.02}{3.4} \):

\[ \frac{1.02}{3.4} \approx 0.3 \]

Thus,

\[ [N_2O_5] \approx 0.3 \times 10^{1} = 3.0 \, \text{mol L}^{-1} \]

### Final Answer

The concentration of \( N_2O_5 \) at that time will be approximately:

\[ [N_2O_5] \approx 3.0 \, \text{mol L}^{-1} \]

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