EXERCISE 1: Kinetic study of urea decomposition In aqueous solution, urea ((NH2)2CO) is likely to decompose into ammonium carbonate according to the reaction: .(H,N), CO+..H,O→..H; +...CO" The reaction rate constant at 350K is 4×10-5 s-1, 5. Compute the time ti required at 350K to decompose 80% of urea. The answer If we decompose 80% of urea, then 0.2 CO of urea is left. Thus In0. 2 = In0.2 In0.2 -kt, i.e. t1 . NA: t1 4x10-5=40236 s= 11,18h k Can you Explain Why?
EXERCISE 1: Kinetic study of urea decomposition In aqueous solution, urea ((NH2)2CO) is likely to decompose into ammonium carbonate according to the reaction: .(H,N), CO+..H,O→..H; +...CO" The reaction rate constant at 350K is 4×10-5 s-1, 5. Compute the time ti required at 350K to decompose 80% of urea. The answer If we decompose 80% of urea, then 0.2 CO of urea is left. Thus In0. 2 = In0.2 In0.2 -kt, i.e. t1 . NA: t1 4x10-5=40236 s= 11,18h k Can you Explain Why?
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![EXERCISE 1: Kinetic study of urea decomposition
In aqueous solution, urea ((NH2)2CO) is likely to decompose into ammonium carbonate
according to the reaction :
.. ..NH; +
(H,N),CO+....H,O→
„Co
2
The reaction rate constant at 350K is 4x10-5 s-1,
5. Compute the time ti required at 350K to decompose 80% of urea.
The answer If we decompose 80% of urea, then 0.2 C0 of urea is left. Thus In0. 2 =
In0.2
-kt, i.e. t1
NA: t1
k
In0.2
==40236 s= 11,18h
-
= -
4x10-5
Can you Explain Why?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F13571319-5b25-421c-ab0c-069a433cb92a%2F878e43dc-4512-4f97-baf6-11c6589fce1a%2F1gsdwt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:EXERCISE 1: Kinetic study of urea decomposition
In aqueous solution, urea ((NH2)2CO) is likely to decompose into ammonium carbonate
according to the reaction :
.. ..NH; +
(H,N),CO+....H,O→
„Co
2
The reaction rate constant at 350K is 4x10-5 s-1,
5. Compute the time ti required at 350K to decompose 80% of urea.
The answer If we decompose 80% of urea, then 0.2 C0 of urea is left. Thus In0. 2 =
In0.2
-kt, i.e. t1
NA: t1
k
In0.2
==40236 s= 11,18h
-
= -
4x10-5
Can you Explain Why?
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