Question
The rate of a first-order reaction is followed by spectroscopy, monitoring the absorption of a colored reactant at 520 nm . The reaction occurs in a 1.29-cm sample cell, and the only colored species in the reaction has an extinction coefficient of 5700 cm-1M-1 at 520 nm .
a) Calculate the initial concentration of the colored reactant if the absorbance is 0.564 at the beginning of the reaction.
b) The absorbance falls to 0.254 at 30.0 min . Calculate the rate constant in units of s-1.
c) Calculate the half-life of the reaction. (in sec)
d) How long does it take for the absorbance to fall to 0.104? (in sec)
a) Calculate the initial concentration of the colored reactant if the absorbance is 0.564 at the beginning of the reaction.
b) The absorbance falls to 0.254 at 30.0 min . Calculate the rate constant in units of s-1.
c) Calculate the half-life of the reaction. (in sec)
d) How long does it take for the absorbance to fall to 0.104? (in sec)
Answers
DrBob222
a) A = ebc
A = 0.564
e = 5700
b = 1.29
c = ?
b) A = ebc
A = 0.254
e = 5700
b = 1.29
c = ? at 30 min.
Convert delta C to delta C/second and rate = k*c
c)
k = 0.693/t<sub>1/2</sub>
You have k, solve for t1/2.
d) A = ebc
A = 0.104
e = 5700
b = 1.29
c = ?
Then ln(No/N) = kt
No = from part a.
N = from part d.
k = from part b.
t = ? in seconds if b is in sec.
A = 0.564
e = 5700
b = 1.29
c = ?
b) A = ebc
A = 0.254
e = 5700
b = 1.29
c = ? at 30 min.
Convert delta C to delta C/second and rate = k*c
c)
k = 0.693/t<sub>1/2</sub>
You have k, solve for t1/2.
d) A = ebc
A = 0.104
e = 5700
b = 1.29
c = ?
Then ln(No/N) = kt
No = from part a.
N = from part d.
k = from part b.
t = ? in seconds if b is in sec.