The reversible chemical reaction A+BC+D has the following equilibrium constant:

Question

The reversible chemical reaction

A+BC+D

has the following equilibrium constant:

Kc=[C][D][A][B]=2.2

Part A:

Initially, only A and B are present, each at 2.00 M. What is the final concentration of A once equilibrium is reached?

Express the molar concentration numerically using two significant figures.

Part B:

What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and[B] = 2.00 M ?

Express the molar concentration numerically using two significant figures.

Explanation / Answer

Part A:
A + B = C + D
2     2      0      0                         ... Initially,
2-x 2-x    x     x                         ... at equilibrium

As Kc = [C][D]/ [A][B] = 2.2

x2/( 2-x )2 = 2.2
On solving the quadratic equation, we get

x= 6.14 and 1.19. As the conc at equilibrium should be less than 2. x= 1.19
[A] = 2-1.19 = 0.81M

Part B :

A + B = C + D
1     2      0      0                         ... Initially,
1-x 2-x    x     x                         ... at equilibrium

As Kc = [C][D]/ [A][B] = 2.2

x2/( 2-x )(1-x) = 2.2
On solving the quadratic equation, we get

x= 4.72 and .78. As the conc at equilibrium should be less than 2. x= 0.78
[D] = 0.78 M

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