The decomposition of sulfuryl chloride (SO2Cl2) is a first-order process. The ra

Question

The decomposition of sulfuryl chloride (SO2Cl2) is a first-order process. The rate constant for the decomposition at 660 K is 4.5x10^-2 ______s-1.

(a) If we begin with an initial SO2Cl2 pressure of 360. torr, what is the pressure of this substance after 66 s? _____torr

(b) At what time will the pressure of SO2Cl2 decline to one fourth its initial value? s By what factor will the pressure of sulfuryl chloride decrease after 6 half lives? Note: The answer wants this value: (360. torr)/value = final pressure after decay______ value

Explanation / Answer

a] For 1st order reaction ;

ln A = - kt + ln[Ao]

ln(P(SO2Cl2)) = -kt + ln(P(SO2Cl2))o
Substituting,
ln(P(SO2Cl2)) = -(4.5 x 10*-2 /s) (66 s) + ln(360 torr)
P(SO2Cl2)) = 18.47 torr

b] The time to arrive at one fourth of the the initial pressure is 2 half lives. For first order reactions the equation for half life is:

t(1/2) = ln(2) / k
t(1/2) = ln(2) / (4.5 x 10*-2 /s) = 15.4 s.

2 half lives time = 30.8 sec

for 1st half life the pressure declines to half of original value

For 6 half lives ====> 1/2^6 = 1/64 of original value

1-1/64 = 63*360/64 = 354.375 torr was the pressure of SO2Cl2

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