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The value of K(p) for the reaction CO(...

The value of `K_(p)` for the reaction
`CO_(2)(g)+C(s)hArr2CO(g)`
is `3.0` bar at `1000 K`. If initially `P_(CO_(2)) = 0.48` bar, `P_(CO) = 0` bar and pure graphite is present then determine equilibrium partial pressue of `CO` and `CO_(2)` . (Multiply answer with 100)

Text Solution

Verified by Experts

The correct Answer is:
15
Let x be the amount of `CO_2` reacted,
`{:(,CO_2(g)+C(s) hArr , 2CO(g)),("Initial pressure","0.48 bar",0),("At equ.","(0.48-x) bar","2xbar"):}`
`K_p=(p_(CO))^2/p_(CO_2)=(2x)^2/(0.48-x)`
or `3=(4x^2)/(0.48-x)`
or `4x^2=3(0.48-x)`
`4x^2=1.44-3x`
or `4x^2 +3x-1.44=0`
Solving for x,
`x=(-3pmsqrt(3^2-4xx4xx(1.44)))/(2xx4)`
`=(-3pm5.66)/8`
`x=2.66/8=0.33`
(negative value of x is neglected because x cannot be negative )
The equilibrium pressures are
`p_(CO)`=2x=2 x 0.33 = 0.66 bar
`p_(CO_2)` = 0.48-x=0.48-0.33=0.15 bar
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