The reaction `2N_(2)O_(5)(g) rarr 4NO_(2)(g)` is first order w.r.t. `N_(2)O_(5)`. Which of the following graphs would yield a straight line?

Options

A. `log p_(N_(2)O_(5))` vs time with `-ve` slope
B. `(p_(N_(2)O_(5)))^(-1)` vs time
C. `p_(N_(2)O_(5)` vs time
D. `log p_(N_(2)O_(5)` vs time with `+ve` slope

Answer

To determine which graph yields a straight line for the reaction \(2N_2O_5(g) \rightarrow 4NO_2(g)\), we need to analyze the reaction's order and the corresponding integrated rate law.

### Step-by-step Solution:

1. **Identify the Order of the Reaction**:
The reaction is given as first order with respect to \(N_2O_5\). This means that the rate of the reaction depends linearly on the concentration of \(N_2O_5\).

2. **Write the Rate Law**:
The rate law for a first-order reaction can be expressed as:
\[ \text{Rate} = k[N_2O_5] \]
where \(k\) is the rate constant.

3. **Integrated Rate Law for First-Order Reactions**:
The integrated rate law for a first-order reaction is given by:
\[ \ln\left(\frac{[A_0]}{[A]}\right) = kt \]
or equivalently,
\[ [A] = [A_0] e^{-kt} \]
where \([A_0]\) is the initial concentration and \([A]\) is the concentration at time \(t\).

4. **Rearranging the Integrated Rate Law**:
We can rearrange the first-order integrated rate law to express it in a linear form:
\[ \ln[A] = -kt + \ln[A_0] \]
This equation is in the form of \(y = mx + c\), where:
- \(y = \ln[A]\)
- \(m = -k\)
- \(x = t\)
- \(c = \ln[A_0]\)

5. **Graphical Representation**:
From the rearranged equation, we can see that plotting \(\ln[A]\) (or \(\ln[N_2O_5]\)) versus time \(t\) will yield a straight line with a slope of \(-k\).

6. **Conclusion**:
Since we are looking for a graph that yields a straight line, the correct graph would be one that plots \(\ln[N_2O_5]\) (or equivalently, \(\log[N_2O_5]\) if using base 10 logarithm) against time.

### Final Answer:
The graph that yields a straight line is \(\ln[N_2O_5]\) versus time (or \(\log[N_2O_5]\) versus time).