the value of Kc = 4.24 at 800K for the reaction
CO(g) + H2O(g) CO2(g) + H2(g)
Calculate the equilibrium concentration of CO2 ,H2O ,CO , H2 at 800K ,if only CO and H2O are pesent initially at concentration of 0.10M each.
For the reaction,
CO (g) + H2O (g) ⇔ CO2 (g) + H2 (g)
Let x mole per litre of each of the productbe formed.
|0.1 - x M||0.1 - x M||x M||x M|
where x is the amount of CO2 and H2 at equilibrium. .
Hence, equilibrium constant can be written as,
Kc= x2/(0.1-x)2 = 4.24
x2 = 4.24(0.01 + x2-0.2x)
x2 = 0.0424 + 4.24x2-0.848x
3.24x2 – 0.848x + 0.0424 = 0
a = 3.24, b = – 0.848, c = 0.0424
for quadratic equation ax2 + bx + c = 0,
Thus solving we get two values of x
x1= 0.067 x2= 0.194
Neglecting x2= 0.194 becuase x could not be more than initial concentration.
Hence the equilibrium concentrations are,
[CO2] = [H2] = x = 0.067 M
[CO] = [H2O] = 0.1 – 0.067 = 0.033 M
0.1 0.1 0 0INITIAL CONCENTRATIONS-AS REACTANTA ARE NOW IN THEIR INITIAL STAGE HAVE JUST STARTED REACTING SO CONC. OF PRODUCTS IS ZERO.NOW,FINAL CONCENTRATIONS.---0.1(1-x) 0.1(1-x) x xconc.reactants conc. productsnow,here we have done 1-x as assuming that from 1 mole of the reactant x amount is converted into products.and since in this case 1 mole of CO2 reactants with 1 mole ofH2O.so equal amounts of both the reactants will decompose to form products.in case of products,co and h2 are in the ratio of 1:1 mole so both will be formed by equal amounts.given Kc=4.24.so,4.24=x*x/(1-x)2hope you can do the remaining calculation......if you like my answer do give a thumbs up.....