1.

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 :

Answer»

`4.60xx10^(-3) s^(-1)`
`6.9xx10^(-3) s^(-1)`
`1.15xx10^(-3) s^(-1)`
`4.90 xx10^(-3) s^(-1)`

SOLUTION :( 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)`


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