The integrated rate laws for zero-, first-, and second-order reaction may be arr

Question

The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line,y=mx+b.

Part A

The reactant concentration in a zero-order reaction was 9.00×102M after 200 s and 2.50×102Mafter 325 s . What is the rate constant for this reaction?

Part B

What was the initial reactant concentration for the reaction described in Part A?

Part C

The reactant concentration in a first-order reaction was 9.20×102M after 25.0 s and 9.90×103Mafter 70.0 s . What is the rate constant for this reaction?

Part D

The reactant concentration in a second-order reaction was 0.300 M after 110 s and 7.70×102M after 845 s . What is the rate constant for this reaction?

Order Integrated Rate Law Graph Slope 0 [A]=kt+[A]0 [A] vs. t k 1 ln[A]=kt+ln[A]0 ln[A] vs. t k 2 1[A]= kt+1[A]0 1[A] vs. t k

Explanation / Answer

*** a ***
for the reaction...
A ---> B + C...

rate = -d[A] / dt = k x [A]^n

where..
[ ] means concentration.
d[A] / dt means change in concentration of A over time
- is because concentration of A is decreasing
k is the rate constant.
[A] is the concentration of A at any given time
n is the order of the reactant and if n = 0, then º means the reaction is zero order in A and zero order overall in this case.

And that equation came from this concept.
rate = change in concentration of A over time...
so..
rate = - d[A] / dt

and..
rate is proportional to the concentration of A at any given point in time. The larger the concentration of A, the faster the rate...

so..
rate is proportional to [A]
and if you put in a proportionality constant
rate = k [A]

therefore...
-d[A] / dt = k x [A]º
for a zero order reaction.

rearranging..
1 / [A]º d[A] = -k dt

and since.. [A]º = 1
(1/1) d[A] = -k dt

integrating from [Ao] to [At].. (initial concentration - which is a constant.. to final concentration which varies by time)
[At] - [Ao] = -k x (t - to)

and if to = 0...ie.. we start the clock at time = 0.. then...
[At] - [Ao] = -k t

rearranging...
[At] = -kt + [Ao]

which is of the form
y = mx + b
if...
y = [At]
m = -k
x = t
b = [Ao]
right?

and since y = mx + b is a line.. a plot of time on the x-axis and concentration of reactant on the y-axis will yield a straight line of slope = -k and intercept = [Ao].. initial concentration.

since you have only 2 data points.. we can just say..
k = - m = - ([A2] - [A1]) / (t2 - t1) = - (2.50x10^-2M - 9*10^-2M) / (325s - 200s) = 5.2x10^-4 M/s

*** B ***
that's the intercept of that line...
[At] = -k x t + [Ao]
[Ao] = [At] + (5.22x10^-4 M/s) x t
pick either point...
[Ao] = ( 9.00×102 M) + (5.2x10^-4 M/s) x (200s) = 0.194M

*** C ***
same deal except..
rate = -d[A] / dt = k x [A]¹

rearranging...
1 / [A]¹ d[A] = -k dt

integrating..
ln[At] - ln[Ao] = -kt

rearranging...
ln[At] = -kt + ln[Ao]

also of the form..
y = mx + b

so a plot of t vs ln[At] gives slope = -k and intercept = ln[Ao]

ie...
k = - (ln[A2] - ln[A1]) / (t2 - t1) = - ln([A2] / [A1]) / (t2 - t1)
k = - ln(9.90×103 / 9.20×102 ) / (70s - 25.0s)
k = 0.0495 / s... notice the units? M/M has cancelled out.

*** d ***
again.. same deal as before...instead of 1 or 0
rate = -d[A] / dt = k x [A]²

rearranging...
1 / [A]² d[A] = -k dt

integrating..
- 1/[At] - -1/[Ao] = -kt

rearranging...
1/[At] -1/[Ao] = kt

rearranging...
1/[At] = kt + 1/[Ao]

also of the form..
y = mx + b

so a plot of t vs 1/[At] gives slope = k and intercept = 1/[Ao]

ie...
k = + (1 / [A2] - 1 / [A1]) / (t2 - t1)
k = + (1/7.70×102M - 1/0.300M) / (845s - 110s)
k = 0.01313 / (Mxsec).

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