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.

The value of K_p for the reaction, CO_(2(g)) + C_((s)) hArr 2CO_((g))

1.	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.
Answer» SOLUTION :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 barand `p_(CO_2)` =(0.48-x)=(0.48-0.3325)=0.1475 bar =0.15 bar