Question

The decomposition of sulfuryl chloride (SO2Cl2) is a first-order process. The rate constant for the decomposition at 660 K is 4.5×10−2s−1. a)If we begin with an initial SO2Cl2 pressure of 460 torr , what is the partial pressure of this substance after 65 s ? b)At what time will the partial pressure of SO2Cl2 decline to one-fifth its initial value?

Answer 1

a) Given that k = 4.5 ×10⁻² s⁻¹

It is understood from the unit of rate constant that it is first order reaction.

For first order reaction,

k =(2.303/t) log (Po/P)

Where ,

k is rate constant

t is the time

Po is the initial pressure

P is the pressure after time t

Given ,

k= 4.5×10⁻²s⁻¹

t = 65 s

Po = 460 torr

Using these values in the above equation, we get

4.5 ×10⁻² s⁻¹= (2.303/65 s ) log(460 torr / P)

On solving for P, we get

P=24.72 torr

Therefore, the partial pressure of substance after 65 s = 24.72 torr

b) Let the Initial pressure = Po

Therefore, the final pressure P= Po/5

time t = ?

The rate constant for first order reaction is given by

k =(2.303/t) log (Po/P)

4.5 ×10⁻² s⁻¹= (2.303/t ) log{Po/ Po/5}

t=[2.303/4.5 ×10⁻² s⁻¹] log 5

t = 35.77 s

Therefore, time required to decline the partial pressure to to one-fifth of its initial value=35.77s