For a homogenous gaseous reaction
`X(g)+2Y)g)hArrZ(g)` ,
at `473 K`, the value of `K_(c )=0.35` concentration units. When `2` moles of `Y` are mixed with `1` mole of `X`, at what pressure `60%` of `X` is converted to `Z`?

Explanation

Since pressure is to be calculated, so first find `K_(p)` using the relation between `K_(c)` and `K_(p)`.
`K_(c)=0.35, R=0.0821, T=473K, Deltan_(g)=1-3=-2`
`K_(p)=K_(c)(RT)^(Deltan_(g))=0.35xx(0.0821xx473)^(-2)=2.32xx10^(-4)`
The expression for `K_(p)` is : `K_(p)=(P_(Z))/(P_(X)(P_(Y))^(2))`

`rArr` Total moles `(n_(r))=3-2x`
Let P = equilibrium pressure
`rArr P_(X)= (1-x)/(3-2x)P, P_(Y)=(2-2x)/(3-2x)P, P_(Z)=(x)/(3-2x)P`
`K_(p)=((x)/(3-2x)P)/((1-x)/(3-2x)P((2-2x)/(3-2x)P)^(2))=(x(3-2x)^(2))/(P^(2)(1-x)(2-2x)^(2))`
`rArr x=0.6` (given)
`K_(p)=(0.6(3-1.2)^(2))/(P^(2)(1-0.6)(2-1.2)^(2))=2.32xx10^(-4) rArr P^(2)= (1.8xx10^(2))^(2)rArr P=180` atm