Question

A spectrophotometer measures changes in absorbance of light over time. Species X has a molar absorptivity constant of 5.00x10^3cmM and the path of the cuvet that holds the mixture is 1.00 cm. Species X, a colored reactant, undergoes a first order decomposition reaction to form a colorless product. See data below.

[X] (M)	absorbance	time (min)
?	0.600	0.0
4.00x10^-5	0.200	35.0
3.00x10^-5	0.150	44.2
1.50x10^-5	0.075	?

a. Calculate the initial concentration of the colored species

b. Using the values given for concentration and time, calculate the rate constant for the first order reaction, include units

c. How many minutes would it take for the absorbance to drop from 0.600 to 0.075

d. What is the half life of the reaction? Include units.

e. The rate constant for this reaction was calculated at various temperatures. Using this data, the graph (to the right) was produced. T stands for temperature.

i.) Label the vertical axis of the graph.

ii.) Explain how you would use this graph to calculate the activation energy.

*Graph is a basic quadrant 1, x-y graph, where the x-axis is labeled (1/T) and the y-axis isn't labeled, a straight line goes down from left to right

Thank you so much for your help, it's an AP Chemistry Ch. 12 FRQ Question and I don't know how to solve it.

Answer 1

from Absorbance (A) A=ebc

e= molar abosrptivity constant= 5*103 M-1 cm-1, b= path length= 1 cm and c=concentration

c= A/eb   hence initial concentration= 0.6/ (5*10³*1) =1.2*10^-4 M

b)

CAO= initial concentration =1.2*10^-4 M

For first order reaction, we known that

lnCA= lnCAO- kt    (1) where CAO= initial concentration and CA= concentration at time t and k is rate constant

from second data point

ln(4*10-5)= ln (CAO)-k*35      (2)

from the third data point

ln(3*10-5)= ln (CAO)- k* 44.2 (3)

Eq. 2- Eq.3 gives

Ln (4/3)= k*(44.2-35) , 0.287682= k*9.2 K= 0.287682/9.2= 0.03127 sec-1

c)

) from Eq.1 ln (CA)= ln CAO- kt

given CA = 1.5*10-5 ( corresponding to 0.075 absorbance and CAO= 1.2*10-4

ln (1.5*10-5)=ln (1.2*10-4)- 0.03127*t

ln (1.5/12)= -0.03127 t

-2.079= -0.03127t and t= time requires =2.079/0.03127=66 sec

d) half life t= 0.693/K for first order reaction or 0.693/0.03127=22.16 seconds

e)