The reversible chemical reaction A+BC+D has the following equilibrium constant:

ID: 977287 • Letter: T

Question

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 the molar concentration numerically using two significant figures.

[A] =

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 the molar concentration numerically using two significant figures.

[A] =

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 the molar concentration numerically using two significant figures.

[D] =

apply

K = [C][D]/[A][B] = 4.6

in equilibrium

[A] = 2-x

[B] = 2-x

[C] = 0+x

[D] = 0+x

then

4.6= [C][D]/[A][B]

4.6 = x*x/(2-x)^2

sqrt(4.6) = x/(2-x)

2.14(2-x) = x

2.14*2 - 2.14x= x

3.14x = 2.14*2

x = 2.14*2/3.14 = 1.363

[A] = 2-1.363 = 0.637

[B] = 2-1.363 = 0.637

[C] = 0+x = 1.363

[D] = 0+x = 1.363

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