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.

If we begin with an initial SO2Cl2 pressure of 450 torr , what is the partial pressure of this substance after 68 s?

At what time will the partial pressure of SO2Cl2 decline to one-third its initial value?

Answer 1

1) Given that k = 4.5 ×10⁻² s⁻¹

It is rate constant of first order reaction because units of first order reaction rate constant = s^-1

For first order recation,

k = 1/t ln { P_o/P_t} -----Eq (1)

Given that

time t = 68 s

Initial pressure P_o = 450 torr

Final pressure P_t = ? torr

k = 1/t ln { P_o/P_t} -----Eq (1)

4.5 ×10⁻² s⁻¹= (1/68 s ) ln {450/ P_t}

P_t = (450)  x e ^{-[4.5×10−2 s−1 x 68 s]}

   = 21.1 torr

Final pressure Pt = 21.1 torr

Therefore, the partial pressure of substance after 68 s = 21.1 torr

2)

k = 1/t ln { P_o/P_t} -----Eq (1)

Initial pressure = P_o

Final pressure P_t = Po/3

time t = ?

k = 1/t ln { P_o/P_t} -----Eq (1)

4.5 ×10⁻² s⁻¹= (1/t ) ln {Po/ Po/3}

t =( In 3 )/ 4.5 ×10⁻² s⁻¹

t = 24.4 s

Therefore,

after 24.4 s , partial pressure of SO2Cl2 decline to one-third its initial value.