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`?

Options

A. 360 atm
B. 180 atm
C. 90 atm
D. 45 atm

Answer

B

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)=2.32 xx 10^(-4)`
The expression for `K_(p)` is : `K_(p)+(p_(z))/(p_(x)(p_(y))^(2))`
`{:("Moles", X, Y, Z),("Initially",1,2,0),("At equilibrium",1-x,2-2x,x):}`
`rArr` total moles `n_(T)=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))`
`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)`
`P=180 atm`