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For a homogenous gaseous reaction X(g)...

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`?

Text Solution

Verified by Experts

Since the pressure is to be calculate, so first find `K_(p)` using the relation between `K_(c )` and `K_(p)`.
`K_(c )=0.35, R=0.0821, T=473`,
For reaction `X(g)+2Y(g)hArrZ(g)`
`Deltan_(g)=n_(P)-n_(R)=1-(2+1)=-2`
`K_(p)=K_(c )(RT)^(Deltan)=0.35xx(0.0821xx473)^(-2)=2.32xx10^(-4)`
[Note: Volume of flask is also not fiven. So, convert `K_(c )` to `K_(p)`]
The expression for `K_(p)` is: `K_(p)=p_(Z)/(p_(X)(p_(Y))^(2))`

rArr Total moles `(n_(T))=3-2x`
Let P= equilibrium pressure, then particle pressure of `X, Y` and `Z` are:
`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"`
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