Using the Equilibrium Constant The reversible chemical reaction A+BC+D has the f

Question

Using the Equilibrium Constant

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

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)                      A   +   B ------------->   C + D         Kc = 4.6

Initial                   2 M     2 M                   0       0

At equilibrium       2-x      2-x                   x        x

                     Kc = x.x / (2-x) (2-x)

                    4.6 = x2/(2-x)2

                           2.145 = x/ 2-x

                 4.290 - 2.145 x = x

                   4.290 = 3.145 x

                      x = 1.364 M

   Therefore,

   equilibrium concentration of A = 2-X = 2 - 1.364 = 0.636 M

B)                           A   +   B ------------->   C + D         Kc = 4.6

Initial                   1 M     2 M                    0       0

At equilibrium       1-x      2-x                  x        x

                     Kc = x.x / (1-x) (2-x)

                    4.6 = x2/ ( 2 - 3x + x2)

                    9.2 - 13.8 x + 4.6 x2 = x2

                            3.6x2 - 13.8 x + 9.2 = 0

                 On solving,

                         x = 0.86 M

Therefore,

final concentration of D at equilibrium = x = 0.86 M

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