Help Part A Exercise 14.52: Problems by Topic - Finding Equilibrium Concentratio

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Part A Exercise 14.52: Problems by Topic - Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Constant What is the equilibrium concentration of CO at 1000 K? For the following reaction, K 255 at 1000 K CO(g) C2(9)COCl2(g) A reaction mixture initially contains a CO concentration of 0.1550 M and a Cl2 concentration of 0.175 M at 1000 K Co0.015 Submit My Answers Give Up Incorrect; Try Again; 2 attempts remaining Part B What is the equilibrium concentration of Cl2 at 1000 K? [C12]= 3.54× 10-2 M Submit My Answers Give Up Correct Part C What is the equilibrium concentration of COCl2 at 1000 K? COC12]= 0.140 M Submit My Answers Give Up Correct

Explanation / Answer

Let's prepare the ICE table

[CO] [Cl2] [COCl2]

initial 0.155 0.175 0

change -1x -1x +1x

equilibrium 0.155-1x 0.175-1x +1x

Equilibrium constant expression is

Kc = [COCl2]/[CO]*[Cl2]

255.0 = (1*x)/((0.155-1*x)(0.175-1*x))

255.0 = (1*x)/(0.02712-0.33*x + 1*x^2)

6.91688-84.15*x + 255*x^2 = 1*x

6.91688-85.15*x + 255*x^2 = 0

Let's solve this quadratic equation

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

a = 2.55*10^2

b = -85.15

c = 6.917

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 = 1.953*10^2

putting value of d, solution can be written as:

x = {85.15 + (1.953*10^2)}/5.1*10^2

x = {85.15 - (1.953*10^2)}/5.1*10^2

solutions are :

x = 0.1944 and x = 0.1396

x can't be 0.1944 as this will make the concentration negative.so,

x = 0.1396

At equilibrium:

[CO] = 0.155-1x = 0.155-1*0.13956 = 0.01544 M

[Cl2] = 0.175-1x = 0.175-1*0.13956 = 0.03544 M

[COCl2] = +1x = +1*0.13956 = 0.13956 M

Answer:

a)

[CO] = 0.0154 M

b)

[Cl2] = 0.0354 M

c)

[COCl2] = 0.140 M

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