Consider the reaction and associated equilibrium constant: aA (g) bB (g), K_c =
ID: 912808 • Letter: C
Question
Consider the reaction and associated equilibrium constant: aA (g) bB (g), K_c = 2.4 Assume that the initial concentration of A is 1.0 M and that no B is present at the beginning of the reaction. Find the equilibrium concentrations of A and B if their a and b coefficients are: a = 1 and b = 1. then for a = 2 and b = 2, and then for a = 2 and b = 1. Consider the reaction: A (g) B (g) + C (g) Find the equilibrium concentrations of A, B, and C for each of the different values of K_c. Assume that the initial concentration of A in each case is 1.0 M and that the reaction mixture initially contains no products. Make any appropriate simplifying assumptions Calculate for K_c= 1.0, then K_c = 1 0 Times 10^-2, then K_c= 1.8 Times 10^-5. (cannot neglect x for first two so must use quadratic for the third you can neglect the x, then x = 0.0042 = [B] and [C], [A] = 1 - 0.0042 = 0.9957 = 1)Explanation / Answer
23)
K = 2.4
aA <-> bB
a)
a = 1 and b = 1
then
A <-> B
Kc = [B]/[A]
initially
[B] = 0
[A] = 1
in equilibrium
[B] = 0 + x
[A] = 1 - x
then
Kc = [B]/[A]
2.4 = (x)/(1-x)
Solve for x
2.4-2.4x = x
3.4x = 2.4
x = 2.4/3.4 = 0.70588
[B] = 0.70588
[A] = 1- 0.70588= 0.29412
for the case
2A <--> 2B
Kc = [B]^2 /[A]^2
similar to the first answer
2.4 = (2x)^2/ (1-2x)^2
solve for x
sqrt(2.4) = 2x/(1-2x)
1.55(1-2x) = 2x
1.55 - 3.10x = 2x
5.10x = 1.55
x =0.303
then
[A] = 1-2x = 1-2*0.303 = 0.394
[B] = 2x = 2*.3030 = 0.606
for last case
2A <-> B
Kc = [B] / [A]^2
2.4 = (x)/(1-2x)^2
2.4(1-4x+4x^2) = x
2.4 - 9.6x + 9.6x^2 = x
9.6x^2 - 10.6x + 2.4 = 0
x = 0.3179 or 0.786 (0.786 will give a negativer answer)
solve for both
[B] = x = 0.3179
[A] = 1-2x = 1-2*0.3179 = 0.3642
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