Chemical equilibrium worksheet A (answer key)



Chemical equilibrium worksheet A (answer key)

1)

Keq=[N2][H2]3 / [NH3]2 = (1.03)(1.62)3 / (0.012)2 = 30410

equilibrium lies right, favors product

2)

PCl5 PCl3 + Cl2

Initial 1.00M 0 M 0 M

Change -X +X +X

(given: X= 13.9% (1.00M)

Equilibrium (1.00-0.139) +0.139 +0.139

Keq=[PCl3][Cl2] / [PCl5] = (0.139)(0.139) / (1.00-0.139) = 0.0224

equilibrium lies left, favors reactants

3)

Keq= 0.00237 M-2 = [NH3]2 / [N2][H3]3 = [NH3]2 / [2.00M][3.00M]3

[NH3]=0.358 M

4)

2NO + O2 2NO2

Initial

Change

Equilibrium X Y X

(This problem initially appears to be an unsolvable two variable equation, until you actually evaluate the Keq)

Keq= 0.00237 M-1 = [NO2]2 / [NO] [O2]2 = X2 / YX2

(The x’s cancel out)

Y= [O2] = 422 M

equilibrium lies right, products favored

5)

decomposition of diatomic iodine to monoatomic iodine

I2 2I

Initial 0.50 M 0

Change -X +2X

Equilibrium (0.50-X) (2X)

Keq= 3.75*10-5 = [I]2 / [I2] = (2X)2 / (0.50-X)

Some problems of this sort require application of the quadratic equation to solve for X, however when Keq is very small (relative to the starting concentrations) X will also be very small, and 0.5-X is close to 0.5.

Keq= 3.75*10-5 = [I]2 / [I2] = (2X)2 / (0.50-X) = 4X2 / (0.5) = 8X2

X=0.002

[I2]=0.5 – 0.002 = 0.5 M (so the assumption of small X was correct)

[I]= 2X= 0.004 M

6)

set up is exactly the same as above, except that initial [I2] is 1.00 M.

decomposition of diatomic iodine to monoatomic iodine

I2 2I

Initial 1.00 M 0

Change -X +2X

Equilibrium (1.00-X) (2X)

Keq= 3.75*10-5 = [I]2 / [I2] = (2X)2 / (1.00-X)

again Keq is small relative to starting concentrations, so X is probably small

Keq= 4X2 = 3.75 * 10-5

X= 0.0031 M

[I2] = 1.00 M (the assumption X ................
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