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A reaction vessel contains 0.100 m cadmium ion, 0.500 m chloride and 0.250 m tetrachlorocadmate ion. which direction will the reaction proceed?

Respuesta :

superman1987
First by considering the following reaction:
  Cd(aq) + 4Cl(aq) ⇄ CdCl4
and according to Kc (the constant of equilibrium ) low:

when Kc = [Concentration of products]^(n) / [ concentration of the reactants]^(n)

when n is the stoichiometric coefficient for the products and reactants
and we have the equilibrium concentrations of:
Cl = 0.5m,  Cd = 0.1 m,    and CdCl4 = 0.250 m 

So by substitution:

∴ Kc = [CdCl4] /( [Cd] [C]^4
        = 0.250 / ( (0.1) * (0.5)^4)
        = 40 
So, ∵ Kc > 1 (that's means the concentration of the products is more than the concentration of the reactants)
-we should have the rate of right reaction equal the rate of left reaction in the equilibrium reaction. so the Kc value should be 1 but when it becomes > 1 that means we need to form more reactants So, that leads the reaction to go leftward to the direction of the reactants to make the equilibrium. 

∴ The reaction goes (leftward) to form more reactants.
   
    

the reaction will be

                            Cd^2+ (aq) +  4Cl^−(aq) ←→  CdCl4^2− (aq)  

iThe Kc of reaction is = 108

Let us calculate Q of reaction

Q = [CdCl4^-2] / [Cd^+2] [ Cl^-]^4

Q = 0.250 / 0.10 X (0.50)^4 = 40

Now as Q < Kc the reaction will go in forward direction or right hand side or towards products


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