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.......

Consider a first order gas phase decomposition reaction given below: A

1.	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.......
Answer» `K=(2.303)/(t)` log `(p_(i))/(p_(i)-x)` `k=(2.303)/(t)` log `(p_(1))/(2p_(i)-p_(t))` `k=(2.303)/(t)` log `(p_(i))/(2p_(i)+p_(t))` `k=(2.303)/(t)` log `(p_(i))/(p_(i)+x)` Solution :Given REACTION :`A_((g))toB_((g))+C_((g))` INITIAL pressure :`p_(i)` 0.00.0 Pressure at t time: `(p_(i)-x)` xx 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