The following data were obtained during the first thermal decomposition of `SO_2Cl_2` at a constant volume, `SO_2Cl_2(g) to SO_2 (g)+Cl_2 (g)` Experiment Time/s Total pressure/ atm 1 0 0.5 2 100 0.6 Calculate the rate of the reaction when total pressure is 0.65 atm.

Explanation

`SO_(2)Cl_(2)(g)rarrSO_(2)(g)+Cl_(2)(g)`
`{:("Initial pressure",p_(0),0,0),("After time t",p_(0)-p,p,p):}`
Total pressure after time `t,(P_(t))=(P_(0)-p)+p+p=P_(0)+p`
or, `p=P_(t)-P_(0)`
Now, `alphapropP_(0) and (a-x)prop(P_(0)-p)`
`therefore(a-x)prop{P_(0)-(P_(t)-P_(0)) }" or, "(a-x)prop2P_(0)-p_(t)`
As the reaction is of the first order,
`k=(2.303)/(t)log.(a)/(a-x)=(2.303)/(t)log.(P_(0))/(2P_(0)-P_(t))`
When t = 100s,
`k=(2.303)/(100)log.(0.5)/(2xx0.5-0.6)=(2.303)/(100)log(2.25)`
`=(2.303)/(100)(0.0969)=2.232xx10^(-3)s^(-1)`
When `P_(t)=0.65"atm",i.e, P_(0)+p=0.65"atm"`
`thereforep=0.65-P_(0)=0.65-0.50=0.15"atm"`
`therefore"Pressure of " SO_(2)Cl_(2) " at time "t (P_(SO_(2)Cl_(2)))`
`=P_(0)-p=0.50-0.15=0.35"atm"`
Rate at that time `kxxp_(SO_(2)Cl_(2))=2.2318xx10^(-3)xx0.35`
`=7.811xx10^(-4)"atm".s^(-1)`