Question

5molee of A are subjected to convert into B according to the given raction: A(g)<=> B (g). At equilibrium 2moles of B is found. Determine;
a) Remaining moles of A at equilibrium.
b) Degree of dissociation of A
c) Mole fraction of A and B at equilibrium
d) Kp and Kc

Answer 1

a)

• Given, A-----> B,
• Total moles of A= 5mol

We know amount of B formed= 2moles

Thus two moles of B were produced from A

Thus 2moles of A has been used up.

Remainig moles of A:

=5-2 mol

= 3mol (ans)

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b)

Degree of dissociation = Amount used up÷ Total amount.

=>Degree of Dissociation of A= 2÷5

=>DD of A=0.4 (ans)

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c)

• let mole fraction of A= Xa
• mole fraction of B = Xb

Xa= Amount of moles left over÷ Total moles of reactant

thus

Xa= 3÷5

=>Xa= 0.6 (ans)

Xb= Amount of moles left÷ Total moles of reactant

Xb= 2÷5

=>Xb=0.4(ans)

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d)

$Kp = \dfrac{[B]}{[A]} \\ \implies \: \dfrac{Pa}{Pb} \\ =>Kp= \dfrac{0.4 \times Pt}{0.6 \times Pt} \\ =>Kp= \dfrac{2}{3}$

$Kc = \dfrac{[B]}{[A]} \\=>Kc = \dfrac{ \dfrac{2}{v} }{ \dfrac{3}{v} } \\ =>Kc= \dfrac{2}{3}$