Part A Initially, only A and B are present, each at 2.00 mol L1. What is the fin
ID: 947479 • Letter: P
Question
Part A
Initially, only A and B are present, each at 2.00 mol L1. What is the final concentration of A once equilibrium is reached?
Express your answer to two significant figures and include the appropriate units.
The reversible chemical reaction
A(aq)+B(aq)C(aq)+D(aq)
Part B
What is the final concentration of Dat equilibrium if the initial concentrations are [A] = 1.00 mol L1 and [B] = 2.00 mol L1 ?
Express your answer to two significant figures and include the appropriate units.
has the following equilibrium constant:
K=[C][D][A][B]=2.5
Explanation / Answer
A)
A(aq)+B(aq)C(aq)+D(aq)
initially:
[A] = 2
[B] = 2
[C] = 0
[D] = 0
in equilibrium
[A] = 2-x
[B] = 2-x
[C] = 0+x
[D] = 0+x
Kc = [C][D]/([A][B])
substitite
Kc = x*x/((2-x)(2-x))
sqrt(Kc) = x/(2-x)
solve for x
2*sqrt(Kc) - x*sqrt(Kc) = x
(1+sqrt(Kc)) *x = 2*sqrt(Kc)
x = 2*sqrt(Kc) / (1+sqrt(Kc))
You will need the value of Kc:
then, get [A]
[A] = 2-x
[A] = 2- 2*sqrt(Kc) / (1+sqrt(Kc))
if Kc = 2.5
[A] = 2- 2*sqrt(2.5) / (1+sqrt(2.5)) = 0.77485 M
B)
A(aq)+B(aq)C(aq)+D(aq)
initially:
[A] = 1
[B] = 2
[C] = 0
[D] = 0
in equilibrium
[A] = 1-x
[B] = 2-x
[C] = 0+x
[D] = 0+x
Kc = [C][D]/([A][B])
substitite
Kc = x*x/((1-x)(2-x))
2.5 = x^2 / (1-3x+x^2)
0.4x^2 = (1-3x+x^2)
0.6x^2 -3x + 1 = 0
x = 0.359127
[A] = 1-x = 1-0.359127 = 0.6408 M