For the first-order reaction A => 3 B + C, when [ A]0 = 0.014 mol·L-1, the concentration of B increases to 0.017 mol·L-1 in 2.7 min.
(a) What is the rate constant for the reaction expressed as the rate of loss of A?
I found this to be .1923, which I know is correct.
(b) How much more time would be needed for the concentration of B to increase to 0.034 mol·L-1?
min
I thought since .034 is twice .017 it would just be doubled...but this is wrong.
ooh nevermind i made a silly error...i have the answer now :)
To find the rate constant for the first-order reaction expressed as the rate of loss of A, you can use the integrated rate law for a first-order reaction.
The integrated rate law for a first-order reaction is:
ln([A] / [A]0) = -kt
Where [A] is the concentration of A at a given time, [A]0 is the initial concentration of A, k is the rate constant, and t is the time.
To solve for the rate constant (k), rearrange the equation as follows:
k = -(1/t) * ln([A] / [A]0)
Now, substitute the given values into the equation. The initial concentration [A]0 is 0.014 mol·L-1, and after 2.7 minutes, the concentration of B increases to 0.017 mol·L-1.
Using these values, we can calculate the rate constant:
k = -(1/2.7) * ln(0.017 / 0.014)
= -0.3704 min^-1
Therefore, the rate constant for the reaction expressed as the rate of loss of A is approximately 0.3704 min^-1.
Now let's move on to part (b). To find how much more time is needed for the concentration of B to increase to 0.034 mol·L-1, we can use the same integrated rate law and rearrange it to solve for time (t).
ln([A] / [A]0) = -kt
Rearranging the equation for time:
t = -(1/k) * ln([A] / [A]0)
Now substitute the given values, where [A] is the final concentration of B (0.034 mol·L-1), and [A]0 is the initial concentration of A (0.014 mol·L-1), and k is the rate constant we found in part (a) (0.3704 min^-1).
t = -(1/0.3704) * ln(0.034 / 0.014)
≈ 5.00 minutes (rounded to two decimal places)
Therefore, approximately 5.00 more minutes would be needed for the concentration of B to increase to 0.034 mol·L-1.