At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO_2 in equilibrium with solid carbon has 90.55% CO by mass. C_((s)) + CO_(2(g)) hArr 2CO_((g)) Calculate K_c for this reaction at the above temperature.

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO_2 in equi

1.	At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO_2 in equilibrium with solid carbon has 90.55% CO by mass. C_((s)) + CO_(2(g)) hArr 2CO_((g)) Calculate K_c for this reaction at the above temperature.
Answer» SOLUTION :Suppose at equilibrium the gaseous mixture is 100 gm, so 90.55% CO in it. 90.55 g CO and (100-90.55) = 9.45 g `CO_2` `{:("Equilibrium reaction :", C_((s))+CO_(2(g)) hArr, 2CO_((g))),("Equilibrium in mass (gm):", 9.45,90.55),("(Mole at equilibrium = Mass/Molecular mass): ",(9.45)/44,(90.55)/28),("(Mole fraction at equilibrium =Mole/Total mole):", "0.2148"/"3.4448","3.234"/"3.4448"):}` Total mole = 0.2148 + 3.234 = 3.4488 `APPROX` 3.45 Partial pressure = Total pressure X Mole fraction `therefore p_(CO_2)` = 1 atm x 0.0624 =0.0624 atm `p_(CO)`=1 atm x 0.9388 =0.9388 atm `K_p=(p_(CO))^2/p_(CO_2)=(0.9388)^2/(0.0624)=14.12` atm `K_p=K_c(RT)^((Deltan(g))` where, `Deltan_((g))` = (2-1) =1 , T= 1127 K , R= 0.0821 atm L `mol^(-1) K^(-1)` `therefore K_c=K_p/(RT)^(Deltan(g))` `=14.12/(0.0821xx1127)^1`=0.1526