Please help Using the Equilibrium Constant Part A The reversible chemical reacti

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Using the Equilibrium Constant Part A The reversible chemical reaction nitially, only A and B are present, each at 2.00 M. What is the final concentration of A once equilibrium is reached? has the following equilibrium constant: Express the molar concentration numerically using two significant figures CID = 6.9 Hints A-0.55 M Submit Kc My Answers Give Up Correct Part B What is the final concentration of D at equilibnum if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M ? Express the molar concentration numerically using two significant figures Hints AS D-3.7 Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining

Explanation / Answer

Let's prepare the ICE table

[A] [B] [C] [D]

initial 1.0 2.0 0 0

change -1x -1x +1x +1x

equilibrium 1.0-1x 2.0-1x +1x +1x

Equilibrium constant expression is

Kc = [C]*[D]/[A]*[B]

6.9 = (1*x)(1*x)/((1-1*x)(2-1*x))

6.9 = (1*x^2)/(2-3*x + 1*x^2)

13.8-20.7*x + 6.9*x^2 = 1*x^2

13.8-20.7*x + 5.9*x^2 = 0

Let's solve this quadratic equation

Comparing it with general form: (ax^2+bx+c=0)

a = 5.9

b = -20.7

c = 13.8

solution of quadratic equation is found by below formula

x = {-b + (b^2-4*a*c)}/2a

x = {-b - (b^2-4*a*c)}/2a

b^2-4*a*c = 1.028*10^2

putting value of d, solution can be written as:

x = {20.7 + (1.028*10^2)}/11.8

x = {20.7 - (1.028*10^2)}/11.8

solutions are :

x = 2.614 and x = 0.895

x can't be 2.614 as this will make the concentration negative.so,

x = 0.895

At equilibrium:

[D] = +1x = 0.895 M

Answer: 0.90 M

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