Question

The decomposition of sulfuryl chloride (so2cl2) is a first-order process. the rate constant for the decomposition at 660 k is 4.5 ✕ 10−2 s-1. (a) if we begin with an initial so2cl2 pressure of 430. torr, what is the pressure of this substance after 70. s?

Answer 1

18.43 Torr is the pressure of this substance after 70seconds.

Step-by-step explanation:

`SO_2Cl_2(g)\rightarrow SO_2(g)+Cl_2(g)`

Rate of the reaction ,k=
`4.5* 10^(-2) s^(-1)`

Integrated rate equation for first order kinetics in gas phase is given as:

`k=(2.303)/(t)\log(p_o)/(p)`

p= pressure of the gas at given time t.

`p_o` = Initial pressure of the gas

When, t = 70 sec

`p_o=430 torr`

`4.5* 10^(-2) s^(-1)=(2.303)/(70 s)\log(p_o)/(p)`

p = 18.43 Torr

18.43 Torr is the pressure of this substance after 70 seconds.

Answer 2

It is a first order process. So the partial pressure at one point is related to partial pressure at other point in time. So the gas-phase reaction is
Ln PSO2Cl2 = Ln (Initial PSO2Cl2) - k x t
Initial Pressure P = 430 = Initial Ln PSO2Cl2.
Time t = 65 s
Constant k = 4.5 âś• 10â’2 s-1
Ln PSO2Cl2 = Ln 430 - (4.5 âś• 10^â’2 x 70)
Ln PSO2Cl2 = 6.063 - 3.15
Ln PSO2Cl2 = 2.913
PSO2Cl2 = 18.43 torr
The pressure of substance after 70 s = 18.43 torr