Question

Suppose 53.96g of COBr2 is placed in a 2.000L flask and heated to 73.0degrees C. Calculate the molar concentrations of each species at equilibrium. and calculate the percentage of the original COBr2 that decomposed at the temp. Kc=0.190. I cant get the quadratic equation right with this. I need to see how you set it up.

Answer 1

Given :

Mass of COBr2 = 53.96 g

Volume of flask = 2.000 L

T = 73.0 ⁰ C = 73.0 ⁰C + 273.15 = 346.15 K

Kc = 0.190

Calculation of concentration of COBr2

Concentration (M) = mol / L

Mol = mass in g / molar mass

[COBr2]= ( 53.96 g / 187.818 g per mol) / 2.000 L

=0.1357 M

Lets show the reaction

COBr2 --- > CO + Br2

I           0.1357             0          0

C         -x                     +x        +x

E          (0.1357-x)       x          x

Kc expression:

Kc = x²/ ( 0.1357 -x) =0.190

X² = 0.190 * ( 0.1357-x)

X² = 0.02728 – 0.190 x

X² + 0.190x – 0.02728 = 0

Lets use quadratic equation formula to get x

x = (-b+sqrt(b²-4ac)) / 2a

Lets write the values of a , b and c

a = 1 , b = 0.190 , c = -0.02728

Lets plug al values to get x

x = ( -0.19 + sqrt ( 0.19²-4 *1 * (-0.02728) ) / 2 *1

x = 0.0955

Now concentration of each species at equilibrium

[COBr2]= 0.1357- x = 0.1357 – 0.0955 = 0.040 M

[Br2]= [CO]= 0.0955 M

Now we have to find the percent dissociation of COBr2

Percent dissociation = (0.0955 / 0.1357) x 100 = 30.0 %