The reaction , `N_2O_4(g)rarr2NO_2(g)` , is first order reaction , which of the following best describes the variation of concentration of `N_2O_4` with time ?

Options

A.
B.
C.
D.

Answer

To solve the problem regarding the variation of concentration of \( N_2O_4 \) with time in the reaction \( N_2O_4(g) \rightarrow 2NO_2(g) \), we need to analyze the characteristics of a first-order reaction.

### Step-by-Step Solution:

1. **Identify the Reaction Order**:
The reaction given is a first-order reaction. This means that the rate of the reaction is directly proportional to the concentration of the reactant \( N_2O_4 \).

2. **Write the Rate Law**:
For a first-order reaction, the rate law can be expressed as:
\[ \text{Rate} = k[N_2O_4] \]
where \( k \) is the rate constant and \( [N_2O_4] \) is the concentration of the reactant.

3. **Use the Integrated Rate Equation**:
The integrated rate equation for a first-order reaction is given by:
\[ \ln\left(\frac{[N_2O_4]_0}{[N_2O_4]}\right) = kt \]
where \( [N_2O_4]_0 \) is the initial concentration at time \( t = 0 \) and \( [N_2O_4] \) is the concentration at time \( t \).

4. **Analyze the Equation**:
From the integrated rate equation, as time \( t \) increases, the natural logarithm of the ratio of the initial concentration to the current concentration increases. This indicates that the concentration of \( N_2O_4 \) decreases over time.

5. **Exponential Decay**:
The concentration of \( N_2O_4 \) decreases exponentially with time. This means that the graph of concentration versus time will show a downward curve that approaches zero but never actually reaches it.

6. **Conclusion**:
Therefore, the variation of concentration of \( N_2O_4 \) with time can be best described as decreasing exponentially.

### Final Answer:
The concentration of \( N_2O_4 \) decreases exponentially with time.

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