The reversible chemical reaction A + B rightwardsharpoonoverleftwardsharpoon C +

Question

The reversible chemical reaction A + B rightwardsharpoonoverleftwardsharpoon C + D has the following equilibrium constant: K_c = [C][D]/[A][B] = 6.5 Initially, only A and B are present, each at 2.00 M. What is the final concentration of A once equilibrium is reached? Express your answer to two significant figures and include the appropriate units. [A] = 0.56 M What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M? Express your answer to two significant figures and include the appropriate units. [D] =

Explanation / Answer

B)

K = [C][D]/([A][B])

6.5 = [C][D]/([A][B])

initially

[A] = 1

[B] = 2

[C] = 0

[D] = 0

in equilibrium

[A] = 1 - x

[B] = 2 - x

[C] = 0+ x

[D] = 0+ x

substitute

6.5 = [C][D]/([A][B])

6.5 = x*x/((1 - x)(2-x))

6.5*(1-3x+x^2) = x^2

6.5-19.5x + 6.5x^2 = x^2

6.5-19.5x + 5.5x^2 = 0

x = 0.3724

[A] = 1 - 0.3724 = 0.6276

[B] = 2 - 0.3724 = 1.6276

[C] = 0+ 0.3724 = 0.3724

[D] = 0+ 0.3724 = 0.3724

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