Consider a first order gas phase decomposition reaction given below:
`A_((g))toB_((g))+C_((g))`
The initial pressure of the system before decomposition of A was `p_(i)` .After lapse of time .t. total pressure of the system increased by x units and became .p_(t)`.The rate constant k for the reaction is given as.......

Options

A. `k=(2.303)/(t)` log `(p_(i))/(p_(i)-x)`
B. `k=(2.303)/(t)` log `(p_(1))/(2p_(i)-p_(t))`
C. `k=(2.303)/(t)` log `(p_(i))/(2p_(i)+p_(t))`
D. `k=(2.303)/(t)` log `(p_(i))/(p_(i)+x)`

Answer

B

Explanation

Given reaction :`A_((g))toB_((g))+C_((g))`
Initial pressure :`p_(i)` 0.0 0.0
Pressure at t time: `(p_(i)-x)` x x
Total pressure =`(p_(i)-x)+(x)+(x)=(p_(i)+x)`
But total pressure =`p_(i)` is given
`therefore p_(t)=(p_(i)+x)` and x=`(p_(t)-p_(i))`
Pressure at t time:
`p_(A)p_(i)-(p_(t)-p_(i))=(2p_(i)-p_(t))=p_(t)=[R]_(t)`
`p_(B)=p_(t)-p_(i)` Initial `p_(A)=p_(i)=[R]_(0)`
`p_(C )=p_(t)-p_(i)`
k of first order reaction =`(2.303)/(t)` log `([R]_(0))/([R]_(t))`
`=(2.303)/(t)` log `((p_(i))/(2p_(x)-p_(t)))`
Thus option (B) is correct