The reversible chemical reaction A+B<-->C+D has the following equilibrium consta

Question

The reversible chemical reaction A+B<-->C+D has the following equilibrium constant: Kc=4.6
What is the final concentration of D at equilibrium if the initial concentrations are [A]=1.00M and [B]=2.00M? The reversible chemical reaction A+B<-->C+D has the following equilibrium constant: Kc=4.6
What is the final concentration of D at equilibrium if the initial concentrations are [A]=1.00M and [B]=2.00M?
What is the final concentration of D at equilibrium if the initial concentrations are [A]=1.00M and [B]=2.00M?

Explanation / Answer

A+B<-->C+D

kc = [C][D]/[A][B]

at equilibrium,

[A] = 1-x

[B] = 2-x

[C] = x , [D] = x

4.6 = x^2/((1-x)(2-x))

x = 0.86 M

at equilibrium , [c] = [D] = 0.86 M

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