In the presence of excess thiocyanate ion, SCN − , the following reaction is first order in chromium(III) ion, Cr 3− ; the rate constant is 2.0 × 10 −6 /s. Cr 3 + ( a q ) + SCN − ( a q ) → Cr ( SCN ) 2 + ( a q ) What is the half-life in hours? How many hours would be required for the initial concentration of Cr 3− to decrease to each of the following values: 25.0% left, 12.5% left, 6.25% left, 3.125% left?
In the presence of excess thiocyanate ion, SCN − , the following reaction is first order in chromium(III) ion, Cr 3− ; the rate constant is 2.0 × 10 −6 /s. Cr 3 + ( a q ) + SCN − ( a q ) → Cr ( SCN ) 2 + ( a q ) What is the half-life in hours? How many hours would be required for the initial concentration of Cr 3− to decrease to each of the following values: 25.0% left, 12.5% left, 6.25% left, 3.125% left?
Author: Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
In the presence of excess thiocyanate ion, SCN−, the following reaction is first order in chromium(III) ion, Cr3−; the rate constant is 2.0 × 10−6/s.
Cr
3
+
(
a
q
)
+
SCN
−
(
a
q
)
→
Cr
(
SCN
)
2
+
(
a
q
)
What is the half-life in hours? How many hours would be required for the initial concentration of Cr3− to decrease to each of the following values: 25.0% left, 12.5% left, 6.25% left, 3.125% left?
Expert Solution & Answer
Interpretation Introduction
Interpretation:
The half-life of SCN- and the time duration for decrease in concentration of SCN- to decrease 25% , 12.5%, 3.125% and 6.5% of its initial value has to be calculated.
Concept Introduction:
Half life period:
The time taken by the concentration of reaction to get reduced of its original concentration is called as half-life reaction.
Half life period can be calculated using the equation,
t1/2=0693k
The half-life period of substance is related to rate constant but it is independent of concentration of reactants.
Answer to Problem 13.65QP
The half life of SCN- is 3.465×105sec(96.25 or 96hr).
The time duration for decrease in concentration of SCN- to decrease 25% of its initial value is 1.9×102hr.
The time duration for decrease in concentration of SCN- to decrease 12.5% of its initial value is 2.9×102hr.
The time duration for decrease in concentration of SCN- to decrease 6.5% of its initial value is 3.9×102hr.
The time duration for decrease in concentration of SCN- to decrease 3.125% of its initial value is 4.8×102hr.
Explanation of Solution
To calculate the half-life of SCN-
Substitute k=9.2/s
Half life period can be calculated using the equation,
t1/2=0693k
t1/2=0.6932.0×10-6/s
t1/2=3.465×105sec(96hr)
The half life of SCN- = 3.465×105sec(96.25 or 96hr).
To calculate time duration for decrease in concentration of SCN- to decrease 25% of its initial value
t25%left=t1/4=2×(96.25hr)=192.5=1.9×102hr
For the concentration to decrease by 25% , it takes 1.9×102hr.
To calculate time duration for decrease in concentration of SCN- to decrease 12.5% of its initial value
t12.5%left=t1/8=3×(96.25hr)=288.75=2.9×102hr
For the concentration to decrease by 12.5% , it takes 2.9×102hr.
To calculate time duration for decrease in concentration of SCN- to decrease 6.5% of its initial value
For the concentration to decrease by 3.125% , it takes 4.8×102hr.
Conclusion
The half-life of N2O5 and the time duration for decrease in concentration of SCN- to decrease to 25% , 12.5%, 6.5% and 3.125% to its initial value was calculated.
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell