The following data pertain to problems 3 and 4. Consider the first-order decompo
ID: 534519 • Letter: T
Question
The following data pertain to problems 3 and 4. Consider the first-order decomposition of hydrogen peroxide at 40 degree C: 2 H_20_2_(aq) + 2 H_2O_(l) + O_2(aq) Kinetic data for the decompositon at 40 degree C are: What is the half-life for the decomposition of hydrogen peroxide at 40 degree C? Express your answer in units of hours, but do not include the units in your submitted answer. If you orignally start with 0.300 M H_2O_2, what will the molarity of H_2O_2 be in the solution after 20.0 hours of reaction? Express your answer units of molarity, but do not include the units in your submitted answer.Explanation / Answer
The integrated rate equation for first order reaction:
K = 2.303/t *log(a/a-x)
Intitial concentration of H2O2 a = 0.300
After 2 hours, a-x = 0.240
Substituting these values in above expression:
K = 2.303/2 *log(0.300/0.240)
K = 1.1515* 0.096910013
K = 0.11159188 /hour
After 4 hour, a-x = 0.189
Substituting these values in above expression:
K = 2.303/4 *log(0.300/0.189)
K = 0.115529679
After 6 hour, a-x = 0.150
Substituting these values in above expression:
K = 2.303/6 *log(0.300/0.150)
K = 0.115545347/ hour
After 8 hour, a-x = 0.120
Substituting these values in above expression:
K = 2.303/8 *log(0.300/0.120)
K = 0.11455698/ hour
Average of rate constant, k = 0.1143 /hour
Half life t1/2 = 0.693/ K
= 0.693/ 0.1143
= 6.06 hour
answer is 6.06
Question 4. After 20 hours rate constant k remains same which is calculated above 0.1143 /hour
K = 2.303/t *log(a /a-x)
0.1143 = 2.303/20 *log(0.300/a-x)
0.1143 = 0.11515 *log(0.300/a-x)
0.992618 = log(0.300/a-x)
0.300/a-x = 10^0.992618
0.300/a-x = 9.831467
a-x = 0.03051 M
thus concentration of H2O2 after 20 hours = 0.03051 M
answer without units = 0.030
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.