The decomposition of IBr(g)into I2 (g) and Br2 (g) is first order in IBr with k = 0.00255/sec.
(a)Starting with [IBr] = 1.50M, what will [IBr] become after 2.50 minutes?
(b)How long, in minutes, will it take for [IBr] to go from 0.500M to 0.100M?
(c)What is the half-life for this reaction in seconds?
(d)Enough IBr is added to an evacuated container to make [IBr] = 0.350M. How long will it take [I2] = 0.100M?
To solve these questions, we can use the integrated rate law for a first-order reaction:
ln([A]t/[A]0) = -kt
where [A]t is the concentration of reactant A at time t, [A]0 is the initial concentration of reactant A, k is the rate constant, and t is the time.
(a) To find [IBr] after 2.50 minutes with an initial concentration of 1.50 M:
We can use the integrated rate law mentioned above. Rearranging the equation, we have:
ln([IBr]/[IBr]0) = -kt
Substituting the given values:
k = 0.00255/sec
[IBr]0 = 1.50 M
t = 2.50 minutes = 2.50 * 60 seconds
We can now solve for [IBr] at time t:
ln([IBr]/1.50) = -(0.00255 sec^-1) * (2.50 min * 60 sec/min)
Simplifying the equation and solving for [IBr]:
[IBr] = e^(-(0.00255 sec^-1) * (2.50 min * 60 sec/min)) * 1.50 M
(b) To find the time it takes for [IBr] to go from 0.500 M to 0.100 M:
Using the same integrated rate law, we can rearrange the equation and set [IBr] = 0.500 M, [IBr]0 = 0.100 M, and solve for time:
ln(0.500/0.100) = -(0.00255 sec^-1) * t
Simplifying the equation and solving for t:
t = ln(0.500/0.100) / -(0.00255 sec^-1)
(c) To find the half-life for this reaction in seconds:
The half-life is the time it takes for the concentration of [IBr] to decrease to half its initial value. The equation for the half-life of a first-order reaction is:
t1/2 = ln(2) / k
Substituting the given value of k, we can solve for t1/2:
t1/2 = ln(2) / 0.00255 sec^-1
(d) To find how long it will take for [I2] to reach 0.100 M when [IBr] is 0.350 M:
Since k is the rate constant for the decomposition of IBr, it is not directly related to the formation of I2. Therefore, we will need additional information to solve this question.