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2. A reaction important in smog formation is. К, - 6.0 х 104 Os(g)NO(g) <=> O2(g) NO2(g) (a) If initial concentrations are [O

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Answer #1

2.

(a)

The reaction quotient

[02] > [NO2 [03] > [NO] 1.5 x 10-4 x 1.5 x 10-4 2.0 10-3 X 2.0 x 10-3 Q = 5.63 x 10-3 Q (5.63 x 10-3)<K (6.0 x 104)

The system is NOT at equilibrium. To reach equilibrium, the reaction will proceed in forward direction. More products will be formed in the forward reaction. This will increase value of Q. It will continue so that Q becomes equal to Kc.

(b)

Let x M be the change in the ozone concentration to reach equilibrium.

Prepare an ICE table.

O_3 NO O_2 NO_2
Initial concentration (M) 2.0 x 10-3 2.0 x 10-3 1.5 x 10-4 1.5 x 10-4
Change in concentration (M) -x -x    +x +x
Equilibrium concentration (M) 2.0 x 10-3-2 2.0 x 10-3-2 1.5 x 10-4+ 1.5 x 10-4+

Substitute values in equilibrium constant expression.

K - (02] [NO21 [03] x [NO] (1.5 x 10-4+) (1.5 x 10-4 +) 6.0 x 109 = (2.0 x 10-3 – 1) x (2.0 x 10-3 –c) 6.0 x 104 = (1.5 x 10-

Take square root on both sides.

(1.5 x 10-4+2) 244.9 = 1 (2.0 x 10-3-1)

244.9(2.0 10-3 – 1) = (1.5 x 10-4 +) 0.48989 – 244.9.x = (1.5 x 10-4 +r) 0.48989 – 1.5 x 10-4 = 244.9.1 +r 0.4897 = 245.9.1

I= 0.00199

The equilibrium concentrations are given below.

103) = NO) = 2.0 x 10-3-1 = 2.0 x 10-3-0.00199 = 8.74 x 10-6 M

[02] = [NO2) = 1.5 x 10-4 + x = 1.5 x 10-4 +0.00199 = 2.14 x 10-3 M

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