The reversible chemical reaction A+B⇌C+D has the following equilibrium constant: Kc=[C][D]/[A][B]=5.4 PART A: Initially, only A...

The reversible chemical reaction A+B⇌C+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!

Solutions

Expert Solution

A+B⇌C+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

queen_honey_blossom answered 1 year ago

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