CHM 152 (Summer 2018) Homework #2 page 3 Integrated Rate Equations and Half-Life
ID: 706522 • Letter: C
Question
CHM 152 (Summer 2018) Homework #2 page 3 Integrated Rate Equations and Half-Life At 25 °C, sucrose reacts with water according to first order kinetics where k - 1.144 x 10 C12H22011 + H20 2C&H1206; rate [C12H201i] 6. If the initial concentration is 0.0500 M, what will be the concentration of sucrose after 30 hours? Hint: the rate constant units are given is inverse seconds 0.0238 M b 0.0145 M c) 0.00832 M d) 0.0397 M e) 00199 M 7. If the initial concentration is 0.0500 M, how many hours will it take for the sucrose concentration to drop to 0.0100M? b) c) 3908 hr 68.42 hr 24.21 hr d) 47.73 hr e) 33.29 hr How many hours does it take for 35 % of the sucrose to react, regardless of its initial concentration? 8, a) 23.9 hr b) 12.2 hr c) 8.37 hr d) 16.2 hr e) 10.5 hr 9. What is the half-life (in hours) at 25 °C? a) 12.75 hr b) 16.83 hr c) 8.497 hr d) 19.46 hr e) 23.15 hr After five half lives, what fraction of sucrose will remain regardless of its initial concentration? 10. a) 1/4 b) 1/16 c) 1/32 d) 1/64 e) 1/128Explanation / Answer
6)
Answer
b) 0.0145M
Explanation
2.303log([A]0/[A]t) = kt
where,
[A]0 = Initial concentration, 0.0500M
[A]t = Concentration at time t,?
k = 1.144×10-5s-1 × 3600 = 4.118×10-2hr-1
t = 30hr
2.303log(0.0500M/[A]t) = 4.118×10-2hr-1 × 30hr
0.0500M/[A]t = 3.4387
[A]t = 0.0145M
7)
Answer
39.08hr
Explanation
t = (2.303/k)log([A]0/[A]t)
= (2.303/4.118×10-2hr-1)log(0.0500M/0.0100M)
= 39.08hr
8)
Answer
e) 10.5hr
Explanation
t = (2.303/k)log([A]0/ [A]t)
= 2.303/4.118×10-2hr-1log (100/65)
= 10.5hr
9)
Answer
b) 16.83hr
Explanation
For first order reaction
t1/2 = 0.693/k
= 0.693/4.118×10-2hr
= 16.83hr
10)
Answer
c) 1/32
Explanation
(1/2)5 = 1/32