The rate constant for a second- order reaction is 0.13 M^-¹s^-¹. If the initial concentration of reactant is 0.26 mol/L, it takes s for the concentration to decrease to 0.13 mol/L.

Question

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The second-order rate law is given as:

rate=k[A]2

We are given k=0.13 MS−1 and [A]0=0.26 M.

We need to find the time for the concentration to decrease to half its initial value, or [A]/2=0.13 M.

Substituting the given values into the integrated rate law for a second-order reaction:

t=1(k[A]0−k[A]t)t[At=0At=[A]/2]

t=1(0.13 M/s)(0.26 M−0.13 M)t=30 s to two significant figures.

Therefore, the answer is choice C) 30.

The Rate Constant for a Second- Order Reaction Is 0