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)`

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

Explanation

The thermal decomposition of `SO_(2)Cl_(2)` at a constant volume
is represented by the following equation.
`SO_(2)Cl_(2(g)toSO_(2(g)+Cl_(2(g)`
`{:("At " t=0,P_(0),0,0),("At " t=t,P_(0)-P,P,P):}`
After time , t, total pressure,
`p_(1)=(P_(1)-p)+P+P`
`rArrP_(1)=P_(0)+P`
`rArrP_=P_(1)+P_(0)`
Therefore, `P_(0)-P=(P_(1)-P_(0))`
=`2P_(0)P_(t)`
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=100s,
`k=(2,303)/100"log"(0.5)/(2xx0.5-0.6)`
`=2.231xx10^(-3) s^(-1)`
When `P_(t) = 0.65` atm,
`rArrP=0.65-P_(0)`
`=.65-0.5`
`0.15`atm
Therefore, when the total pressure is 0.65 atm, pressure of
`SOCl_(2)` is
`P_(SOCl_(2)=P_(0)_P`
`= 0.5-0.15`
`=0.35`atm
Therefore, the rate of equation, when total pressure is 0.65
atm, is given by,
`"Rate=k^((P_(SOCl_(2))))`
`=(2.23xx10^(-3) s^(-1))(0.35 atm)`
`=7.5xx10^(-4)"atm" s^(-1)`