Question

initially co2 and C are taken in 1mole each in a closed vessel The total P is 2atm calculate KP ( NOTE) at eq 0.1mol of Co is formed Co² + C equilibrium 2co what is kp​

Answer 1

Answer:

+ CO2(g) + heat = H2O(l) + CO(g) balanced eq

1 : 1 : 1 : 1 ratio

Keq is the equilibrium constant at given temperature. Keq = [C] × [D] / [A] × [B] This equation is called equation of law of chemical equilibrium. At equilibrium, the concentration of reactants is expressed as moles/lit so Keq = Kc and if it expressed as partial pressure then Keq = Kp.

V = 5L

Initially 0.8mole/5L = 0.16mole/L = {H2} = {CO2}

Keq = ({H2O} x {CO} / ({CO2} x {H2}) = x^2 / (0.16-x)

Keq = 4 = x^2 / (0.16-x) and x^2 = 0.64 -4x

x^2 + 4x = 0.64 and x = 0.154mole/L

{CO2} = 0.16 - 0.154 = 0.006mole/L

Explanation:

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