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]=5.4

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.

Thanks!

Explanation / Answer

A+BC+D

For th chemical reaction, given Kc = [C] [D] /[A] [B] =5.4

              A    +     B         C   +    D

Initially    2.00     2.00         0         0

Change   -x          -x            x    x

At equilibium 2.00-x     2.00-x      x           x

Kc = 5.4 = x.x / (2.00-x) (2.00-x) = x2 / (2.00-x)2

5.4 (2.00-x)2 = x2

-4.4x2 - 21.6x + 21.6 = 0

- [ x2 + 4.9x - 4.9] = 0

x2 + 4.9x - 4.9 = 0

x = 0.710M

[A] = 2 - 0.710 = 1.28 M

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.