Question

The decomposition of sulfuryl chloride (SO₂Cl₂) is a first order process. The rate constant at 660 K is 4.5 x 10^-2s^-1.

a) If the initial pressure of sulfuryl chloride is 375 torr, what is the pressure of the substance after 65 sec?

b) At what time will the pressure of sulfuryl chloride decline to one tenth of its original value?

Answer 1

solution a) The reaction is first order reaction.

Integrated 1st order equation : ln(p) = ln(p₀) -kt

on rearranging the above equation : ln (p/p₀) = -kt

where, p is pressure of sulfuryl chloride and p₀ is its initial pressure.

p₀ = 375torr

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

t = 65 sec

By substituting values in equation : ln (p/p₀) = -kt

ln(p/375) = -(4.5 * 10^-2s^-1) (65sec)

ln(p/375) = -2.925

solving the above equation we get, p = 20.12 torr

pressure after 65 sec will be 20.12 torr

b) we have to find the time when pressure of sulfuryl chloride decline to one tenth of its original value.

initial pressure(p₀) = 375

one tenth of initial pressure (p)= 375/10 = 3.75

we can find the time form equation : ln(p) = ln(p₀) -kt

on rearranging: ln(p/p₀) =-kt

ln(32.5/325) = -(4.5 * 10^-2s^-1) (t)

solving the above equation for t, we get t = 51.16sec

At 51.16sec the pressure of sulfuryl chloride decline to one tenth of its original value.