Question

- As described in Exercise 14.43, the decomposition of sulfuryl chloride (SO₂Cl₂) 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 pressure of this substance after 60 s?

(b) At what time will the pressure of SO₂Cl₂ decline to one-tenth its initial value?

Answer 1

use integrated rate law for 1st order reaction

A = A0e^-Kt

given
rate constant K = 4.5 × 10–2 s–1
time t= 60 s
Initial pressure A0 = 450 torr

final pressure A= 450 torr * e^- 4.5 × 10–2 s–1* 60 s
= 30.2 Torr
rounding to 2 significant figures
a) Answer : 30. Torr
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use integrated rate law for 1st order reaction

A = A0e^-Kt

ln A= lnA0 -Kt

Kt = ln{A0/A]

A= 1/10 A0 = 1/10 * 450 torr = 45 torr

rate constant K = 4.5 × 10–2 s–1

kt = ln{A0/A]
4.5 × 10–2 s–1 *t= ln [450/45]

t= ln10/4.5 × 10–2 s–1
= 51.2 sec
Rounding to 2 significant figures

Answer : 51 S
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hope it helps !