The reversible chemical reaction A+B?C+D has the following equilibrium constant:

Question

The reversible chemical reaction

A+B?C+D

has the following equilibrium constant:

Kc=4.8

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] = 1.00M and [B] = 2.00M ?

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

Explanation / Answer

To do this you'll need to set up a ICE table (or as I use a RICES table)

Reaction:----A--+---B---<=>--C---+---D
Initial:--------1.00---1.00--------0--...
Change:-----(-x)----(-x)---------(+x)-...
Equilibrium:-(1-x)--(1-x)-------(x)---...
Stop:--------(FIND)-------------------...

[In table D and C are Initial = 0, Change = +x, Equilibrium = x, Stop is blank, Yahoo cut off my table :( ]

We want to find A at Stop. To do this we use the values at equilibrium and plug them into the equilibrium equation:

[C][D]/[A][B] = Kc
[x][x]/[1-x][1-x] = 4.8

x^2 / (1-x)^2 = 4.8 Now take the square root of both sides
x/ (1-x) = 2.19 Cross multiply
x = 2.19 - 2.19x Move x over
3.19x = 2.19 Solve for x
x = 0.687



(1- 0.687) = 0.313 M of A at equilibrium

Now do the same for the second part except the I values will be different and you are looking for D this time.

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