The reversible chemical reaction A + B C + D has the following equilibrium const

Question

The reversible chemical reaction A + B C + D has the following equilibrium constant: K_c = [C][D]/[A][B] = 4.6 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. 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.

Explanation / Answer

Part A

at eqlibrium

[A] = [B] = 2.00-x

[C] = [D] =x

4.6 = x^2/(2.00-x)^2

square root

2.14 = x/2-x

4.28 - 2.14x = x

x= 1.36 M

[ A ] = 2.00-1.36 = 0.64 M

Part B

the reversible chemical reaction

A + B C + D

1...2...0...0

1-x...2-x...x...x

so k = 4.6 = x^2/(1-x)*(2-x)

x = D = 1.4 M

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