Question

If 32.5 % of a reactant in a certain first-order reaction decomposes 540.0 seconds after initiation of the reaction, what is the half-life?

a) 245 s
b) 333 s
c) 878 s
d) 952 s

Answer 1

The half-life of a first-order reaction is calculated using the integrated rate equation and the relationship t1/2 = 0.693/k. By determining the rate constant k from the given 32.5% decomposition over 540 seconds, we can solve for the half-life.

Step-by-step explanation:

To calculate the half-life of a first-order reaction when 32.5% of the reactant decomposes in 540 seconds, we can use the first-order integrated rate equation:

ln([A]0/[A]) = kt

Where:

[A]0 is the initial concentration of the reactant,

[A] is the concentration of the reactant at time t,

k is the first-order rate constant, and

t is the time.

Since 32.5% decomposes, 67.5% remains, which can be represented as a fraction (0.675) of the initial concentration. We can rearrange the equation to solve for k, and then use the relationship of the half-life (t1/2) for a first-order reaction:

t1/2 = 0.693/k

Assuming we have determined k from the given 32.5% decomposition:

ln(1/0.675) = k * 540

Now we solve for k:

k = ln(1/0.675) / 540

Once k is determined, we calculate the half-life:

t1/2 = 0.693/k

By plugging the determined value of k into this equation, we can find the half-life that corresponds to one of the multiple-choice answers provided.