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_(i))`
D. `k = (2.303)/(t) log (p_(i))/(p_(i) + x)`

Answer

B

Explanation

If x atm be the decrease in pressure of A at time t and one mole each of B and C is being forward, the increase in pressure of B and C will also be x atm each.
`{:(.,A_((g)) to,B_((g)) +,C_((g))),(At t = 0,p_(i) atm,0 atm,0atm),("At time" t, (p_(i) - x) atm, x atm,.):}`
Where `p_(i)` is the initial pressure at time t = 0
`p_(t) = (p_(i) - x) + x + x = p_(i) + x`
`x = (p_(t) - p_(i))`
where, `P_(A) = P_(i) - x = P_(i) - (P_(t) - P_(i)) = 2P_(i) - P_(i)`
`k = ((2.303)/(t)) ("log"(P_(i))/(P_(A))) = (2.303)/(t) "log" (P_(i))/((2P_(i) - P_(i))`