For a first-order rate law, which plot is a straight line?

For a first-order rate law, which plot is a straight line?

• A. Concentration versus time
• B. Ln concentration versus time
• C. 1 over concentration versus time
• D. Temperature versus time

Answer: (B)

A first-order rate law produces exponential decay of the reactant concentration over time. Writing the differential equation d[A]/dt = -k[A] and separating variables gives d[A]/[A] = -k dt. Integrating leads to ln[A] = -kt + ln[A]0, which is a straight line when you plot ln[A] versus time, with slope -k and intercept ln[A]0. This linear relationship is why the natural logarithm of the concentration plotted against time is the correct straight-line plot for first-order kinetics. Plotting concentration versus time shows a curved, exponential decrease, not a straight line. Plotting 1/[A] versus time would be linear for a second-order process, not first-order. Plotting temperature versus time doesn’t reflect the direct time dependence of concentration in the rate law.

A first-order rate law produces exponential decay of the reactant concentration over time. Writing the differential equation d[A]/dt = -k[A] and separating variables gives d[A]/[A] = -k dt. Integrating leads to ln[A] = -kt + ln[A]0, which is a straight line when you plot ln[A] versus time, with slope -k and intercept ln[A]0. This linear relationship is why the natural logarithm of the concentration plotted against time is the correct straight-line plot for first-order kinetics.

Plotting concentration versus time shows a curved, exponential decrease, not a straight line. Plotting 1/[A] versus time would be linear for a second-order process, not first-order. Plotting temperature versus time doesn’t reflect the direct time dependence of concentration in the rate law.