Question 1

Why does a first-order reaction remove equal fractions rather than equal numbers of molecules?

Accepted Answer

In a first-order reaction, the rate is proportional to the current concentration: -d[A]/dt = k[A]. During every equal interval of duration delta-t, the remaining fraction is e^(-k delta-t), so the fraction lost is constant even though the absolute number lost gets smaller.

Question 2

What happens to the reactant after successive first-order half-lives?

Accepted Answer

The fraction remaining after n half-lives is (1/2)^n. Thus, the remaining fractions are 50%, 25%, 12.5% and 6.25% after one, two, three and four half-lives. Each half-life takes the same amount of time.

Question 3

Why is a plot of ln[A] against time a straight line for a first-order reaction?

Accepted Answer

The integrated rate law is ln[A] = -kt + ln[A]0. This has the form y = mx + c, with y = ln[A], slope m = -k and intercept c = ln[A]0.

Question 4

How does a common-log plot differ from a natural-log plot?

Accepted Answer

For log base 10, the integrated equation uses 2.303: k = (2.303/t) log([A]0/[A]). Therefore, a graph of log[A] against t is also straight, but its slope is -k/2.303 rather than -k.

Question 5

Does a zero-order reaction have the same half-life formula?

Accepted Answer

No. For a zero-order reaction, [A]t = [A]0 - kt, or equivalently the amount reacted is x = kt. Its half-life is t1/2 = [A]0/(2k), so it depends on the initial concentration. The constant half-life t1/2 = 0.693/k is the first-order case.

First-Order Decay, Half-Life & Log Plots in 3D

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