CHM 152 (Summer 2018) Homework #2 page 3 Integrated Rate Equations and Half-Life

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) 1/128

Explanation / 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

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