The graph which represents the following reaction is:

The given reaction is a unimolecular (single-reactant) decomposition, for example $$N_2O_5 \rightarrow 2\,NO_2 + \tfrac12\,O_2$$. Such decompositions follow first-order kinetics.

Step 1: Write the differential rate law for a first-order reaction.
$$-\frac{d[A]}{dt}=k[A]$$

Step 2: Integrate the rate law.
Rearrange and integrate between $$t = 0$$ and any time $$t$$:
$$\int_{[A]_0}^{[A]} \frac{d[A]}{[A]} = -k\int_{0}^{t} dt$$
$$\ln[A] - \ln[A]_0 = -kt$$
or
$$\ln[A] = -kt + \ln[A]_0 \qquad -(1)$$

Step 3: Interpret the integrated form.
Equation $$(1)$$ is of the form $$y = mx + c$$ with
$$y \equiv \ln[A], \; x \equiv t, \; m = -k \;(\text{negative slope}), \; c = \ln[A]_0 \;(\text{intercept}).$$
Hence a plot of $$\ln[A]$$ versus $$t$$ must be a straight line with a negative slope.

Step 4: Match with the given options.
Among the four graphs, Option A is the only one showing a straight line for $$\ln[A]$$ (or $$\log[A]$$) versus time with a downward slope. Therefore, Option A correctly represents the kinetics of the reaction.

Option A which is: the straight-line plot of $$\ln[A]$$ versus $$t$$ (negative slope).