Reactant A undergoes dissociation into products B and C according to the followi

Question

Reactant A undergoes dissociation into products B and C according to the following equilibrium reaction equation

A rigid vessel, initially containing 0.075 mol/L of pure A, is allowed to equilibrate at 105°C.

a) What would be the value of the equilibrium constant (Keq expressed in terms of the molar concentrations) if the equilibrium concentrations of all species are equal?
(No units required.)

b) What would be the value of the equilibrium constant (Keq expressed in terms of the molar concentrations) if, at equilibrium, the concentration of A is 1.0% that of B?

(No units required.)

Explanation / Answer

The reaction is A (g)<---> B(g)+C(g)

let x= moles of A dissociated to reach equilibrium

At Equilibrium A =0.075-x and B= x and C=x

but given at equilibrium, all the concentrations are equal

0.075-x=x

2x= 0.075 and x= 0.075/2=0.0375

KC= Equilibrium constant = [C] [B]/ [A]= 0.0375

b) given the concentration of A is 1% of B

0.075-x =(1/100)*x

0.075-x =0.01x, x=0.0743

KC= [0.0743*0.0743/ 0.007)= 0.788

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