? Chapter 14 Problem 14.43 9of22 Part A The decomposition of suituryl chloride (

Question

? Chapter 14 Problem 14.43 9of22 Part A The decomposition of suituryl chloride (SO Cl2) is a frst-order process. The rate constant for the decompositon at 660 K is 4.5x10-18-1. If we begin with an initial S02Clz pressure of 470 torr, what is the partial pressure of this substance after 86 s? Express your answer using two significant figures torr X incorrect; Try Again; 4 attempts remaining ?PartB Al what time will the parsial pressune of 50 Cl2 decine to one-fourth its initial value? Express your answer using two significant figures 2 3 4 5 6 9

Explanation / Answer

Ans 14.43

Rate constant k = 4.5*10^-2 s-1

Part a

For first order reaction

ln [P(SO?Cl?)0 / P(SO?Cl?)] = kt

Initial pressure

P(SO?Cl?)? = 470 torr

Time t = 66 s

Final pressure P(SO?Cl?) = ?

ln [470 / P(SO?Cl?)] = 4.5*10^-2 x 66

470 / P(SO?Cl?) = exp (4.5*10^-2 x 66)

470 / P(SO?Cl?) = 19.49

P(SO?Cl?) = 24.112 torr

Part b

Given that

P(SO?Cl?) = (1/4) P(SO?Cl?)0

P(SO?Cl?)0 / P(SO?Cl?) = 4

For first order reaction

ln [P(SO?Cl?)? / P(SO?Cl?)] = (4.5 × 10?² s?¹) t

ln(4) = (4.5 × 10?² s?¹) t

t = ln(4) / (4.5 × 10?² s?¹)

t = 30.81 s

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