Using Graphs to Determine Rate Part A The reactant concentration in a zero-order

Question

Using Graphs to Determine Rate

Part A

The reactant concentration in a zero-order reaction was 0.100M after 175s and 1.00

Using Graphs to Determine Rate

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. 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

Part A

The reactant concentration in a zero-order reaction was 0.100M after 175s and 1.00

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
-ve 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]
[At] - [Ao] = -k * (t - to)
if to = 0
[At] - [Ao] = -kt
[At] = -kt + [Ao]

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

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
k = - m = - ([A2] - [A1]) / (t2 - t1) = -(1.00 x 10-2 M - 0.100M) / (310s - 175s)
k = 6.67 * 10-4 M/s

(b)
that's the intercept of that line
[At] = -kt + [Ao]
[Ao] = [At] + (6.67 * 10-4 M/s) x t
pick either point
[Ao] = (0.100M) + (6.67 * 10-4 M/s) x (175s)
[Ao] = 0.217 M

(c)
same process as above
rate = -d[A] / dt = k x [A]1

rearranging
1 / [A]1 d[A] = -k dt

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

k = - (ln[A2] - ln[A1]) / (t2 - t1) = - ln([A2] / [A1]) / (t2 - t1)
k = - ln(6.50 x 10^-3 / 5.10x10^-2) / (60s - 30s)
k = 0.069 s-1

(d)
Again same process
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]

k = +(1 / [A2] - 1 / [A1]) / (t2 - t1)
k = +(1/2.20x10^-2M - 1/0.510M) / (790s - 235s)
k = 0.0784 M-1sec-1.

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