Monitoring First-Order Decomposition of Species X with Spectrophotometer

[disneychannel]

The first-order decomposition of a colored chemical species, X, into colorless products is monitored with a spectrophotometer by measuring changes in absorbance over time. Species X has a molar absorptivity constant of 5.00 x 10^3 cm^ -1 M^ -1 and the path length of the cuvette containing the reaction micture is 1.00 cm. The data from the experiment are given in the table below.
[X] Absorbance Time(min)
(M)

? 0.600 0.0
4.00 x 10^ -5 0.200 35.0
3.00 x 10^ -5 0.150 44.2
1.5 x 10^ -5 0.075 ?

a) calculate the initial concentration of the colored species.
- I got 1.2 * 10 ^ -4 is that right?

b) calculate the rate constant for the first-order reaction using the values given for concentration and time. Include unites with your answer.
- I got 0.0314, however I do not know if this is correct and will the units be cm ^ -1 M ^ -1

c) calulate the number of minutes it takes for the absorbace to drop from 0.600 to 0.075
how do you do this?
d) calculate half-life of the reation. Include units with your answer.
e) http://www.collegeboard.com/prod_downloads/ap/students/chemistry/ap-cd-chem-0607.pdf
question # 3

I WOULD REALLY APPRECIATE if you help me out, it is really confusing!

THANKS!

 

[Zefram]

Hi there,

I am happy to help you with your questions! Let's go through them one by one:

a) To calculate the initial concentration of the colored species, we can use the Beer-Lambert Law, which relates the absorbance of a solution to its concentration. The equation is A = εcl, where A is the absorbance, ε is the molar absorptivity constant, c is the concentration, and l is the path length. Rearranging the equation, we get c = A/(εl). Plugging in the given values, we get c = 0.600/(5.00 x 10^3 cm^-1 M^-1 * 1.00 cm) = 1.2 x 10^-4 M. So your answer is correct!

b) To calculate the rate constant, we can use the integrated rate law for a first-order reaction, which is ln([X]t/[X]0) = -kt, where [X]t is the concentration at a given time, [X]0 is the initial concentration, k is the rate constant, and t is time. Rearranging the equation, we get k = -ln([X]t/[X]0)/t. Plugging in the values from the table, we get k = -ln(1.5 x 10^-5 M/1.2 x 10^-4 M)/44.2 min = 0.0313 min^-1. Your answer is correct, and the units for the rate constant are indeed min^-1.

c) To calculate the time it takes for the absorbance to drop from 0.600 to 0.075, we can use the equation ln(A/A0) = -kt, where A is the absorbance at a given time, A0 is the initial absorbance, k is the rate constant, and t is time. Rearranging the equation, we get t = -ln(A/A0)/k. Plugging in the values, we get t = -ln(0.075/0.600)/0.0313 min^-1 = 19.2 min. So it takes 19.2 minutes for the absorbance to drop from 0.600 to 0.075.

d) The half-life of a reaction is the time it takes for half of the reactant to be consumed.

 

[quicknote]

a) To calculate the initial concentration of the colored species, we can use the Beer-Lambert Law, which relates the absorbance of a solution to its concentration and the path length of the cuvette. We can rearrange the equation to solve for concentration:

C = A / (ε * d)

Where:
C = concentration (M)
A = absorbance
ε = molar absorptivity constant (cm^-1 M^-1)
d = path length (cm)

Plugging in the values given in the table, we get:

C = 0.600 / (5.00 x 10^3 * 1.00)
C = 1.2 x 10^-4 M

So, your calculation is correct.

b) To calculate the rate constant, we can use the integrated rate law for a first-order reaction:

ln([X]t / [X]0) = -kt

Where:
[X]t = concentration at time t (M)
[X]0 = initial concentration (M)
k = rate constant (s^-1)
t = time (s)

We can rearrange the equation to solve for k:

k = -ln([X]t / [X]0) / t

Plugging in the values given in the table, we get:

k = -ln((0.200 / 1.2 x 10^-4) / 35.0)
k = 0.0314 s^-1

So, your calculation is also correct. The units for the rate constant are s^-1, not cm^-1 M^-1.

c) To calculate the time it takes for the absorbance to drop from 0.600 to 0.075, we can use the integrated rate law again:

ln([X]t / [X]0) = -kt

We can rearrange the equation to solve for t:

t = -ln([X]t / [X]0) / k

Plugging in the values given in the table, we get:

t = -ln((0.075 / 0.600) / 0.0314)
t = 20.6 min

So, it takes approximately 20.6 minutes for the absorbance to drop from 0.600 to 0.075.

d) The half-life of a first-order reaction can be calculated using the half-life equation:

t1/2