The value of `K_p` for the reaction,
`CO_(2(g)) + C_((s)) hArr 2CO_((g))` is 3.0 at 1000 K. If initially `p_(CO_2)` = 0.48 bar and `p_(CO)` = 0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and `CO_2`.

Explanation

Suppose, the decrease in concentration (pressure) of `CO_2` in reaction = x bar. So, according to stoichiometry of reaction, the increase in pressure of `CO_2` is x bar.
`{:("Reaction equilibrium:",CO_(2(g))+ , C_((s)) hArr , 2CO_((g))),("Initial pressure :","0.48 bar",-,0),("Change of pressure :","-x bar",-,+ 2x "bar"),("At equilibrium :", (0.48-x)"bar",,2x "bar"):}`
For this chemical equilibrium , equilibrium constant `K_p`,
`therefore K_p=(P^2CO)/(PCO^2)`
where , `K_p`=3.0
`p_(CO)`=2x bar
`p_(CO_2)`=(0.48-x) bar
`therefore 3.0=(2x)^2/((0.48-x))`
`therefore 4x^2=3(0.48-x)=1.44-3x`
`therefore 4x^2+3x-1.44=0`
`x=(-b pm sqrt(b^2-4ac))/(2a)`
So, a=4 , b=3 , c=-1.44
`therefore x=(-3pmsqrt((3)^2-4(4)(-1.44)))/(2(4))`
`=(-3pmsqrt(9+23.04))/8=(-3pmsqrt(32.04))/8`
`=(-3pm5.660)/8`
`=2.66/8` OR `(-8.66)/8`
=0.3325 OR -1.0825
The value of x cannot be negative .
`therefore` x=0.3325 bar
So, `p_(CO)`=2x=2(0.3325)=0.6650 bar and `p_(CO_2)` =(0.48-x)=(0.48-0.3325)=0.1475 bar
=0.15 bar