The reversible chemical reaction
A+B⇌C+D has the following equilibrium constant:Kc=[C][D][A][B]=3.3 |
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.
SubmitHintsMy AnswersGive UpReview Part 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. |
K = C*D/(A*B)
K = 3.3
A = 2
B = 2
C = 0
D = 0
find A in equilibrium
A = 2-x
B = 2 -x
C = 0+x
D = 0+x
substitute all in
K = C*D/(A*B)
3.3= (x*x)/(2-x)^2
3.3 = x^2 /(2-x)^2
solve
sqrt(3.3)= x/(2-x)
1.8165 = x/(2-x)
0.55050x = 2-x
1.55050x = 2
x = 2/1.55050 = 1.289906
since
A = 2-x= 2-1.289906 = 0.710094 M of A
pretty similar
A = 1-x (this changes!)
B = 2 -x
C = 0+x
D = 0+x
substitute all in
K = C*D/(A*B)
3.3= (x*x)/(2-x)(1-x)
3.3 = x^2 /(2-3x+x^2)
solve
(2-3x+x^2) = 0.303*x^2
0.697 x^2 -3x +2 = 0
x = 0.8246
subsitute in d
D = 0+x = 0.8246 M
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