A gaseous reaction A(g) `rarr 2B (g) +C(g)` is found to be of first order . If the reaction is started with `p_(A)` =90 mm Hg the total pressure after 10 min is found to be 180 min Hg. The rate constant of the reaction is :

Options

A. `4.60xx10^(-3) s^(-1)`
B. `6.9xx10^(-3) s^(-1)`
C. `1.15xx10^(-3) s^(-1)`
D. `4.90 xx10^(-3) s^(-1)`

Answer

C

Explanation

( C) ` A(g) rarr 2B (g) + C(g) `
After 10 min , `p_(0)-p 2p p `
Pressure after 10 min `P_(0)-p+2p+p=p_(0) +2p`
180 mm Hg =90 mm Hg +2P
or 2p= 90 mm mg
`:. P` = 45 mm Hg
Now `k = (2.303)/(t) "log" (p_(0))/(P)`
`=(2.303)/(10 "min") "log" (90)/(45)`
`= (2.303)/(10xx60s)xx0.303`
`= 1.155xx10^(-3) s^(-1)`