Following is a first order gas phase reaction: A_((g))toB_((g))+C_((g)).At t time ,total pressure =p_(t) and partial pressure of A=p_(A) atm.So,derive the integrated rate equation for this reaction .

Following is a first order gas phase reaction: A_((g))toB_((g))+C_((g)

1.	Following is a first order gas phase reaction: A_((g))toB_((g))+C_((g)).At t time ,total pressure =p_(t) and partial pressure of A=p_(A) atm.So,derive the integrated rate equation for this reaction .
Answer» <P> Solution :Suppose,the partial pressure of A,B and C are `p_(A),P_(B)` and `p_(c)` respectively so,`p_(t)=p_(A)+p_(B)+p_(C)` Ifx atm be the decrease in pressure of A at time t and one MOLE each of B and C is being formed,the increase in pressure of B and C will also be x atm each. These expression is in following equation So,total pressure (`p_(i)`-x+x+x)=`p_(t)` `therefore p_(i)+x=p_(t)` `therefore x=(p_(i)-p_(i)` atm) Thus at equilibrium `p_(A)=(p_(i)-x)` atm `(2p_(i)-p_(t))=[R]_(t)` At initially t=0 time `p_(i)=[R]_(0)` For FIRST order reaction `k=(2.303)/(t)` LOG `([R]_(0))/([R]_(t))` `therefore k=(2.303)/(t)` log `(p_(i))/((2p_(i)-p_(i)))`