The dissociation equilibrium of a gas $$AB2$$ can be represented $$as2AB2g$$ ⇌ $$2AB(g)$$ $$+$$ $$B2(g)The$$ degree of dissociation is x and is small compared to $$1$$. The expression relating to the degree of dissociation (x) with equilibrium constant Kp and total pressure P is

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The dissociation equilibrium of a gas $$AB2$$ can be represented $$as2AB2g$$  ⇌   $$2AB(g)$$  $$+$$   $$B2(g)The$$ degree of dissociation is x and is small compared to $$1$$. The expression relating to the degree of dissociation (x) with equilibrium constant Kp and total pressure P is

Infinity Learn · Accepted Answer

$$2AB2g$$  ⇌   $$2AB(g)$$  $$+$$   $$B2(g)Initial$$ moles              $$1$$                      $$0$$                 $$0$$                    At equilibrium     $$21-x$$                 $$2x$$                 xTotal moles at equilibrium $$=$$ $$2$$ $$-$$ $$2x$$ $$+$$ $$2x$$ $$+$$ x $$=$$ $$(2$$ $$+$$ $$x)So$$,          $$pAB2$$ $$=$$  $$21-xp2+x$$,   pAB $$=$$  $$2xp2+x$$      $$pAB2$$ $$=xp2+xKp$$ $$=$$ $$pAB2$$  $$pAB2pAB22=2xp2+x2x2+xp$$ $$21-xp2+x$$ $$=$$ $$x3p2$$ $$+$$ x  $$1-x2$$  $$=$$   $$x3p2$$ ∵    x  <<  $$1$$  and $$2$$  So,  $$1-x$$  ≈  $$1$$  and $$2+x$$  ≈  $$2$$. x $$=$$ $$2Kpp1/3$$

The dissociation equilibrium of a gas AB2 can be represented as2AB2g ⇌ 2AB(g) + B2(g)The degree of dissociation is x and is small compared to 1. The expression relating to the degree of dissociation (x) with equilibrium constant Kp and total pressure P is

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