Consider a first order gas phase decomposition reaction given below:
`A(g) to B_(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_(i))/(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

Let us consider a first order gas phase decomposition reaction:
`A(g) to B(g)+C(g)`
The initial pressure of the system before decomposition of A is `.p_(I).`. After lapse of time .t.. Total pressure of the system increased by x units and became `P_(t).`.
Hence, the pressure of A decreassed by x atom.
Initial pressure: `P_(i)` atom 0 0
Pressure after time t: `(P_(i)-x)" "x" "atm" "x" "atm`
`p_(i)=(P_(i)-x)+x+x`
`=P_(i)+x` atm.
`x_(i)=P_(t)-P_(i)`
`=P_(i)-(P_(t)+P_(i))`
`P_(A)=2P_(i)-P_(t)`
`k=(2.303)/(t)"log"([A]_(0))/([A])`
`=(2.303)/(t)"log"(P_(i))/(2P_(i)-P_(t))`