Consider the data given below for hypothetical reaction `A rarr X`
`{:(,"Time (s)",,"Rate of the reaction"(mol L^(-1)s^(-1))),(,0,,1.60xx10^(-2)),(,10,,1.60xx10^(-2)),(,20,,1.60xx10^(-2)),(,30,,1.59xx10^(-2)):}`
From the above data, the order of reaction is:

Options

A. Zero
B. 1
C. 2
D. 3

Answer

To determine the order of the reaction from the provided data, we will analyze the relationship between the rate of the reaction and the concentration of the reactant over time.

### Step-by-Step Solution:

1. **Understanding the Data**:
- We have the following data:
- At time 0 seconds, the rate of reaction is \(1.60 \times 10^{-2}\) mol L\(^{-1}\) s\(^{-1}\).
- At time 10 seconds, the rate is still \(1.60 \times 10^{-2}\) mol L\(^{-1}\) s\(^{-1}\).
- At time 20 seconds, the rate remains \(1.60 \times 10^{-2}\) mol L\(^{-1}\) s\(^{-1}\).
- At time 30 seconds, the rate is \(1.59 \times 10^{-2}\) mol L\(^{-1}\) s\(^{-1}\), which is approximately the same as the previous rates.

2. **Analyzing the Rates**:
- The rate of the reaction remains approximately constant (around \(1.60 \times 10^{-2}\) mol L\(^{-1}\) s\(^{-1}\)) despite the passage of time. This indicates that the rate does not change significantly as the concentration of reactant A decreases.

3. **Understanding Reaction Order**:
- The general rate equation for a reaction is given by:
\[ \text{Rate} = k \cdot [A]^n \]
where \(k\) is the rate constant, \([A]\) is the concentration of reactant A, and \(n\) is the order of the reaction.

4. **Determining the Order**:
- Since the rate of reaction remains constant even as the concentration of A decreases, we can infer that the rate does not depend on the concentration of A.
- This implies that \(n = 0\) because if the concentration of A does not affect the rate, the reaction behaves as a zero-order reaction.

5. **Conclusion**:
- Therefore, based on the data provided, the order of the reaction is 0.

### Final Answer:
The order of the reaction is **0** (Option A).

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