Which plot yields a straight line for a zero-order rate law?

Which plot yields a straight line for a zero-order rate law?

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

Answer: (A)

For zero-order kinetics, the rate is a constant, not dependent on the concentration: rate = k. This means the reactant concentration decreases at a constant amount each unit time. Mathematically, -d[A]/dt = k, and integrating gives [A] = [A]0 − kt. So [A] vs time is a straight line with slope −k and y-intercept [A]0. That linear relationship is why plotting concentration against time yields a straight line for a zero-order process. If you plot the natural logarithm of concentration versus time, you get ln([A]0 − kt), which is not a straight line. Plotting 1/[A] versus time gives 1/[A] = 1/([A]0 − kt), also not linear. Temperature versus time isn’t tied to the zero-order concentration change in a way that guarantees a straight line unless temperature—and thus k—changes in a specific, coordinated way.

For zero-order kinetics, the rate is a constant, not dependent on the concentration: rate = k. This means the reactant concentration decreases at a constant amount each unit time. Mathematically, -d[A]/dt = k, and integrating gives [A] = [A]0 − kt. So [A] vs time is a straight line with slope −k and y-intercept [A]0.

That linear relationship is why plotting concentration against time yields a straight line for a zero-order process. If you plot the natural logarithm of concentration versus time, you get ln([A]0 − kt), which is not a straight line. Plotting 1/[A] versus time gives 1/[A] = 1/([A]0 − kt), also not linear. Temperature versus time isn’t tied to the zero-order concentration change in a way that guarantees a straight line unless temperature—and thus k—changes in a specific, coordinated way.