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)` .
A
`P_(CO)=0.15bar, P_(CO_(2))= 0.66 bar`
B
`P_(CO)=0.66 bar, P_(CO_(2))= 0.15 bar`
C
`P_(CO)=0.33 bar, P_(CO_(2))= 0.66 bar`
D
`P_(CO)= 0.66 bar, P_(CO_(2)) = 0.33 bar`
Text Solution
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The correct Answer is:
B
The value of `K_(p)` for the reaction ………….. `{:(,CO_(2)(g)+,C(s)hArr2,CO(g)),(t=0,0.48,-,0),(t=equi^(m),0.48-x,-,2x):}` `K_(p) = ((2x)^(2))/((0.48-x)) = 3` `x = 0.33 bar`
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