For a first-order reaction $$\mathrm{R}\rightarrow\mathrm{P}$$ at a given temperature, $$k$$ is the rate constant. For this reaction, at the given temperature, the concentrations of $$\mathrm{R}$$ and $$\mathrm{P}$$ at a time $$t$$ are $$[\mathrm{R}]$$ and $$[\mathrm{P}]$$, respectively. The correct graphical representation(s) for this reaction is(are)

For a first-order reaction $$\mathrm R \rightarrow \mathrm P$$ at constant temperature we have

$$\text{rate}=k[\mathrm R]$$

and the integrated rate-law is

$$\ln[\mathrm R]=\ln[\mathrm R]_0-k\,t \qquad-(1)$$

where $$[\mathrm R]_0$$ is the initial concentration of R.

Equation $$(1)$$ is of the straight-line form $$y=mx+c$$ with

• ordinate $$y=\ln[\mathrm R]$$
• abscissa $$x=t$$
• slope $$m=-k$$ (negative)
• intercept $$c=\ln[\mathrm R]_0$$.

Hence a plot of $$\ln[\mathrm R]$$ versus $$t$$ must be a straight line having a negative slope equal to $$-k$$. This is the characteristic graphical test for first-order kinetics.

Among the four alternatives the only diagram that shows a straight line for $$\ln[\mathrm R]$$ vs $$t$$ with negative slope is Option A. Every other option either plots the wrong variable, gives a curve, or shows the wrong sign of slope.

Therefore the correct graphical representation is given by:

Option A which is: $$\ln[\mathrm R]\; \text{(ordinate)} \; \text{vs}\; t\; \text{(abscissa)}$$ — a straight line with slope $$-k$$.