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]=7.5

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 your answer to two significant figures and include the appropriate units.

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 your answer to two significant figures and include the appropriate units.

Explanation / Answer

a)

Answer

0.54M

Explanation

A + B <------> C + D

K = [ C ] [ D ]/[ A ] [ B ]= 7.5

at equillibrium

[ A ] = 2.00 - x

[ B ] = 2.00 - x

[ C ] = x

[ D ] = x

Therefore,

x^2/(2.00 - x)^2 = 7.5

taking square root on both side

x/(2.00 - x ) = 2.74

3.74x = 5.48

x = 1.465

Therefore, at equillibrium

[ A ] = 2.00 - 1.465 = 0.54M

b)

Answer

0.90M

Explanation

At equillibrium

[ A ] = 1.00 - x

[ B ] = 2.00 - x

[ C ] = x

[ D ] = x

therefore,

x^2/(1.00 - x ) (2.00 - x ) = 7.5

x = 0.9013

Therefore,

[ D] = 0.90M

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