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)toSO_(2) (g) +Cl_(2)(g)`

Calculate the rate of reaction when total pressure is `0.65` atm.

Explanation

`SO_(2)Cl_(2) (g)toSO_(2) (g) +Cl_(2)(g)`

Total pressure after time t
`i.e., p_(t)=p_(i) -p+p+p=p_(i)+p`
So,` a=p_(i)`
`a-X =p_(i) -(p_(t)-p_(i))p_(i)-p_(t) +p_(i)=2p_(i)-p_(t)`
Substitutions of the value of a and `(a-x)` given
`k=(2.303)/(t)log ""(p_(i))/((2p_(i)-p_(t)))=(2.303)/((100s))log ""(0.5 atm)/(0.4 atm)`
`=(2.303)/((100s))log 1.25 =(2.303)/((100s))xx0.0969`
`=2.23xx10^(-3)s^(-1)`
b) Calculation of reaction rate when total pressure is `0.65` atm
`P_(So_(2)Cl_(2))=0.5 -(0.65-0.50)=(1-0.65)=0.35` atm
`k= 2.23 xx10^(-3)s^(-1)`
Rate `=kxxP _(SO_(2)Cl_(2))=(2.23xx10^(-3)s^(-1))xx(0.35atm)`
Rate `=7.8 xx10^(-4) atm s^(-1).`