For a second-order reaction, how does the half-life change as concentration decreases?

For a second-order reaction, how does the half-life change as concentration decreases?

• A. It increases
• B. It decreases
• C. It stays constant
• D. It becomes negative

Answer: (A)

In a second-order reaction that depends on a single reactant, the rate law is -d[A]/dt = k[A]^2. When you integrate, you get t = (1/k)(1/[A] - 1/[A]0). To find the half-life from the initial concentration to half of that amount, set [A] = [A]0/2. This gives t1/2 = 1/(k[A]0). So the half-life is inversely proportional to the starting concentration: as [A]0 decreases, t1/2 increases. More generally, the time required to reduce the current concentration by half is t1/2' = 1/(k[A]), which also grows as [A] decreases. That means the half-life gets longer as the concentration falls.

In a second-order reaction that depends on a single reactant, the rate law is -d[A]/dt = k[A]^2. When you integrate, you get t = (1/k)(1/[A] - 1/[A]0). To find the half-life from the initial concentration to half of that amount, set [A] = [A]0/2. This gives t1/2 = 1/(k[A]0). So the half-life is inversely proportional to the starting concentration: as [A]0 decreases, t1/2 increases. More generally, the time required to reduce the current concentration by half is t1/2' = 1/(k[A]), which also grows as [A] decreases. That means the half-life gets longer as the concentration falls.