A first-order reaction plots linearly as:

For a first-order reaction, the rate of reaction is proportional to the concentration of one reactant. The differential rate law is expressed as: $$-\frac{d[A]}{dt} = k[A]$$ To find the relationship between concentration and time, we rearrange and integrate this expression: $$\frac{d[A]}{[A]} = -k dt$$ Integrating both sides from $t=0$ to $t=t$: $$\int_{[A]_0}^{[A]_t} \frac{1}{[A]} d[A] = -k \int_{0}^{t} dt$$ $$\ln[A]_t - \ln[A]_0 = -kt$$ This yields the integrated rate law for a first-order reaction: $$\ln[A]_t = -kt + \ln[A]_0$$ Comparing this to the equation for a straight line ($y = mx + b$): - $y = \ln[A]_t$ - $x = t$ - $m$ (slope) $= -k$ - $b$ (y-intercept) $= \ln[A]_0$ Therefore, a plot of $\ln[A]$ versus $t$ will be a straight line with a slope of $-k$.