Using the Equilibrium Constant A+B?C+D K c=[C][D][A][B]=5.9 Part A Initially, on

Question

Using the Equilibrium Constant

A+B?C+D

Kc=[C][D][A][B]=5.9

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 the molar concentration numerically using two significant figures.

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Part B

What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00M and [B] = 2.00M ?

Express the molar concentration numerically using two significant figures.

Using the Equilibrium Constant

The reversible chemical reaction

A+B?C+D

has the following equilibrium constant:

Kc=[C][D][A][B]=5.9

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 the molar concentration numerically using two significant figures.

[A]=   M

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Part B

What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00M and [B] = 2.00M ?

Express the molar concentration numerically using two significant figures.

[D]=   M

Explanation / Answer

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

Kc = {[C]*[D]}/{[A]*[B]} = 5.9

Now, initially [A] = [B] = 2 M

Let at eqb, [A] = [B] = (2-x)M

Thus, at eqb. [C] = [D] = x M

Thus, Kc = x2/(2-x)2 = 5.9

or, x/(2-x) = sqrt(5.9) = 2.43

or, x = 1.42 M

thus at eqb. [A] = [B] = 2-X = 0.58 M

[C] = [D] = x = 1.42 M

2) If initially [A] = 1 M ; [B] = 2 M

Let at eqb. [A] = (1-x) & [B] = (2-x)M

thus, at eqb. [C] = [D] = x M

now, Kc = x2/{(2-x)*(1-x)} = 5.9

or, x2 = 5.9x2 - 17.7x + 11.8

or, 4.9x2 - 17.7x + 11.8 = 0

or, x = 0.88 M

Thus, at eqb. [D] = x = 0.88 M

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