# (a) What is meant by homogeneous equilibrium? Give an example. (b) The value of Kc = 4.24 at 800 K for reaction.

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(a) What is meant by homogeneous equilibrium? Give an example.

(b) The value of Kc = 4.24 at 800 K for reaction.

CO(g) + H2O(g) ⇌ CO2 (g) + H2(g)

Calculate equlibrium concentration of CO2, H2, CO and H2O at 800K, if only CO and H2O are present initially at concentrations of 0.10 M each.

(C) For which equlibrium, the vapour pressure is constant at a given temperature.

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Homogeneous Equilibrium: Equilibrium in a system having only one phase is called homogeneous equilibrium.

e. g. N2 (g) + 3H2 (g) ⇌ 2NH3(g)

(b) For the reaction, Let x mole L−1 of each the product be formed

At equilibrium (0.1- x) M (0.1 - x) M xM xM where x = amount of CO2 and H2 at equlibrium.

Hence, equilibrium constant can be written as –

Kc$\frac{X^2}{(0.1-X)^2}$ = 4.24

x2 = 4.24(0.01 + x20.2x)

x2 =  0.0424 + 4.25x2 − 0.848x)

3.24x2 − 0848x ± 0.848, c = 0.0424

For quadratic equation ax2 + bx + c = 0

x = ${(-b \pm \sqrt{b^2-4ac}) \over 2a}$

x = ${(0.848 \pm \sqrt{(0.848)^2-4\times3.24\times0.0424}) \over 2\times3.24}$

$\frac{0.848 \pm 0.4118}{6.48}$

x1$\frac{0.848 \pm 0.4118}{6.48}$ = 0.067

x2$\frac{0.848 \pm 0.4118}{6.48}$ = 0.194

The value 0.194 should be neglected because it will give concentration of the reactant which is more than initial concentration. There, the equilibrium concentrations ate –

[CO2] = [H2] = x = 0.067 M

= [CO] = [H2O] = 0.1 - x

= 0.1 - 0.067

= 0.033 M

(c) For liquid ⇌ vapour equilibrium, the vapour pressure is constant at a given temperature.