Solution Found!
As described in Exercise 14.41, the decomposition of
Chapter , Problem 14.43(choose chapter or problem)
As described in Exercise 14.41, the decomposition of sulfuryl chloride 1SO2Cl22 is a first-order process. The rate constant for the decomposition at 660 K is 4.5 * 10-2 s -1 . (a) If we begin with an initial SO2Cl2 pressure of 450 torr, what is the partial pressure of this substance after 60 s? (b) At what time will the partial pressure of SO2Cl2 decline to one-tenth its initial value?
Questions & Answers
QUESTION:
As described in Exercise 14.41, the decomposition of sulfuryl chloride 1SO2Cl22 is a first-order process. The rate constant for the decomposition at 660 K is 4.5 * 10-2 s -1 . (a) If we begin with an initial SO2Cl2 pressure of 450 torr, what is the partial pressure of this substance after 60 s? (b) At what time will the partial pressure of SO2Cl2 decline to one-tenth its initial value?
ANSWER:Problem 14.43
As described in Exercise 14.41, the decomposition of sulfuryl chloride 1SO2Cl22 is a first-order process. The rate constant for the decomposition at 660 K is 4.5 * 10-2 s -1 . (a) If we begin with an initial SO2Cl2 pressure of 450 torr, what is the partial pressure of this substance after 60 s? (b) At what time will the partial pressure of SO2Cl2 decline to one-tenth its initial value?
Step by Step Solution
Step 1 of 2
a)
The given decomposition reaction of sulfuryl chloride is as follows.
From the given,
Rate constant “k” =
Pressure of the reactant = 450 torr
Time “t” = 60 s
According to the first order reaction,
The integral form of the rate equation is as follows.
Substitute the given values in the above equation as follows.
Therefore, the partial pressure of the substance is 30 torr.
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