For the reaction `2N_(2)O_(5)(g)to4NO_(2)(g)+O_(2)(g)`, the following results have been obtained :
`{:("S. No.",,,[N_(2)O_(5)]" mol L"^(-1),,,"Rate of disappearance of"),(,,,,,,N_(2)O_(5)","" mol L"^(-1)min^(-1)),(1,,,1.13xx10^(-2),,,34xx10^(-5)),(2,,,0.84xx10^(-2),,,25xx10^(-5)),(3,,,0.62xx10^(-2),,,18xx10^(-5)):}`
(a) Calculate the order of reaction
(b) Write rate law
(c) Calculate rate constant of the reaction.

Explanation

To solve the problem step by step, we will analyze the given reaction and the data provided.

### Step 1: Write the balanced chemical equation
The balanced reaction is:
\[ 2N_{2}O_{5}(g) \rightarrow 4NO_{2}(g) + O_{2}(g) \]

### Step 2: Identify the rate law
The rate law for a reaction can generally be expressed as:
\[ \text{Rate} = k [N_{2}O_{5}]^n \]
where \( k \) is the rate constant, \( [N_{2}O_{5}] \) is the concentration of the reactant, and \( n \) is the order of the reaction.

### Step 3: Use the provided data to find the order of the reaction
We have the following data:

| S. No. | [N2O5] (mol L^-1) | Rate of disappearance of N2O5 (mol L^-1 min^-1) |
|--------|-------------------|--------------------------------------------------|
| 1 | 1.13 x 10^-2 | 34 x 10^-5 |
| 2 | 0.84 x 10^-2 | 25 x 10^-5 |
| 3 | 0.62 x 10^-2 | 18 x 10^-5 |

To find the order of the reaction, we can compare the rates and concentrations from two different trials.

Using the first two trials:
- Trial 1:
- Concentration: \( [N_{2}O_{5}]_1 = 1.13 \times 10^{-2} \)
- Rate: \( R_1 = 34 \times 10^{-5} \)

- Trial 2:
- Concentration: \( [N_{2}O_{5}]_2 = 0.84 \times 10^{-2} \)
- Rate: \( R_2 = 25 \times 10^{-5} \)

Using the rate ratio:
\[ \frac{R_1}{R_2} = \frac{k [N_{2}O_{5}]_1^n}{k [N_{2}O_{5}]_2^n} \implies \frac{34 \times 10^{-5}}{25 \times 10^{-5}} = \left(\frac{1.13 \times 10^{-2}}{0.84 \times 10^{-2}}\right)^n \]

Calculating the left side:
\[ \frac{34}{25} = 1.36 \]

Calculating the right side:
\[ \frac{1.13}{0.84} \approx 1.48 \]

Thus, we have:
\[ 1.36 = (1.48)^n \]

Taking logarithm on both sides to solve for \( n \):
\[ \log(1.36) = n \cdot \log(1.48) \]
Calculating the logarithms:
\[ n = \frac{\log(1.36)}{\log(1.48)} \approx 0.85 \approx 1 \text{ (approximately)} \]

### Step 4: Write the rate law
Since we found \( n \approx 1 \), the rate law can be written as:
\[ \text{Rate} = k [N_{2}O_{5}]^1 = k [N_{2}O_{5}] \]

### Step 5: Calculate the rate constant \( k \)
We can use any of the trials to calculate \( k \). Using trial 1:
\[ k = \frac{R_1}{[N_{2}O_{5}]_1} = \frac{34 \times 10^{-5}}{1.13 \times 10^{-2}} \]
Calculating \( k \):
\[ k = \frac{34 \times 10^{-5}}{1.13 \times 10^{-2}} \approx 0.03008 \text{ min}^{-1} \]

### Final Answers
(a) The order of the reaction is approximately **1**.
(b) The rate law is **Rate = k [N2O5]**.
(c) The rate constant \( k \) is approximately **0.03008 min^{-1}**.