Question

For the reaction `A_(2)(g) + 2B_(2)hArr2C_(2)(g)` the partial pressure of `A_(2)` and `B_(2)` at equilibrium are `0.80` atm and `0.40` atm respectively.The pressure of the system is `2.80` atm. The equilibrium constant `K_(p)` will be
A. (A) `20`
B. (B) `5.0`
C. (C) `0.02`
D. (D) `0.02`

Answer 1

Correct Answer - A
`A_(2)(g)+2B_(2)(g)hArr2C_(2)(g)`
`P_(A_(2)=0.80atm. P_(B_(2)=0.4` atm.
Total pressure of the system`=2.8` atm.
`thereforeP_(C_(2)=2.8-0.8-0.4=1.6`
`Kp=(P_(C_(2)^(2))/(P_(A_(2)+P_(B_(2)^(3))))=((1.6)^(2))/(0.8+(0.4))^(2)=20`