Question

The following data were obtained during the first order thermal decomposition of `SO_(2) Cl_(2)` at a constant volume :
`SO_(2)Cl_(2) (g) to SO_(2)(g)+Cl_(2)(g)`
`{:("Experiment Times"//s^(-1) " Total pressure/atm"),("1 0 0.4" ),("2 100 0.7"):}`
Calculate the rate constant. Given : log 4 = 0.6021, log 2 = 0.3010)

Answer 1

The thermal decomposition of `SO_(2) Cl_(2)` at a constant volume is represented by the following equation :
`SO_(2)Cl_(2) (g) to SO_(2)(g)+Cl_(2)(g)`
`{:("At t" = 0, P , O, O),("At t"= 0, P_(0) - P, P, P):}`
`P_(1) = (P_(0) - P) + P + P`
After time t, total pressure, `P_(t) = P_(0) + P`
Which on rearrangement gives : `p = _(t) - P_(0)`
Therefore, `P_(0) - P = P_(0) - (Pt - P_(0))= 2P_(0) - P_(1)`
For a first -order reaction,
`k = (2.303)/(t) "log" (P_(0))/(P_(0) - P)`
` = (2.303)/(t) "log" (P_(0))/(2P_(0) - P_(t))`
When t = 100 s,
` k= (2.303)/(100) "log" (0.4)/(2xx0.4- 0.7)(2.303)/(100)"log"(0.4)/(0.1)`
`rArr = (2.303)/(100) log^(4) =(2.303)/(100)xx0.6021 = 0.01387S^(-1)`