When a system is at equilibrium a) the forward and reverse reaction rates are equal. Observable macroscopic properties do not change 2a) This reaction is exothermic, so heat is on the product side. Reducing temperature will make it shift to the product side maximizing the production of NO. 2b) Decreasing pressure by increasing volume will favour the product side since there are 10 molecules on the right and 9 on the left.

2c A platinum catalyst would have no effect since it speeds up both forward and reverse reactions equally. 2d Rate equations depend on concentration and adding He would increase pressure but not change the concentration so there would be no effect.

3. H2(g) + I2(g)  2HI(g) mol/L Initial 4.00 mol/2.00L 2.00 Shift @E 0.40 mol/2.00L 0.20 +0.20 +0.20 -0.40 1.60 0.20 Ke = 1.602 / (0.20)2 = 64

4. 2NO(g) + O2(g)  2NO2(g) mol/L Initial Shift @E x 0.606 0.190 6.45 x 105= (0.190)2/ (0.606)x2 (6.45 x 105)(0.606)x2 = (0.190)2 x2 = (0.190)2 / [(6.45 x 105)(0.606)] x = 3.04 x 10-4

5. H2(g) + CO2(g)  H2O(g) + CO(g) mol/L Initial Shift @E 12.0/20.0 0.600 8.0/20.0 0.400 10.0/20.0 0.500 x 4.40 = (0.500)x/ (0.400)(0.600) x = (4.40) (0.400)(0.600) / 0.500 x = 2.11 mol/L x 20.0 L = 42.2 mol

6. 2A(s) + 3B(g)  2C(g) + D(s) Remember in a heterogeneous system solids are not included in the equilibrium constant Ke = [C]2 / [B]3 1.00 x 10-2 = [C]2 / 0.203 [C]2 = (1.00 x 10-2) (0.203) [C] = 8.90 x 10-3 mol/L

7. HCHO(g) <==> H2(g) + CO(g) mol/L Initial 4.00 mol/4.00L 1.00 Shift @E -0.20 +0.20 +0.20 0.80 0.20 0.20 Ke =(0.20)2 / 0.80 Ke = 0.050 mol/L

8. H2(g) + I2(g)  2HI(g) mol/L Initial 1.00 1.00 Shift @E -x -x +2x 1.00 - x 1.00 - x 2x 0.22 0.22 1.6 49.0 = (2x)2 / (1.00 – x)2 7.0 = 2x / 1.00 – x 7.00 – 7x = 2x 9x = 7 x = 7/9 = 0.78

9. SO2(g) + NO2(g)  SO3(g) + NO(g) mol/L Initial 0.60 0.20 0.80 0.30 Upset x Shift @E 0.50 -0.20 -0.20 +0.20 +0.20 0.40 x 1.00 (0.80)(0.30)/(0.20)(0.60) = (0.50)(1.00)/(x)(0.40) x = (0.20)(0.60)(0.50) / (0.40)(0.80)(0.30) x =0.625 mol/L x 2.0 L = 1.25 mol

10. H2(g) + I2(g)  2HI(g) mol/L Initial 0.10 0.20 0.60 Upset Shift @E 2.00 -x -x +2x 2.1 - x 0.20 - x 0.60 + 2x Ke =(0.60)2 / (0.20)(0.10) = 18 18 = (0.60 + 2x)2 / (2.1-x)(0.20-x) 18 = 0.36 + 2.4 x + 4x2 / 0.42-2.1x-0.20x+x2 7.56-41.4x+18x2 = 0.36 + 2.4 x + 4x2 14x2-43.8x+7.2=0 Use x = -b+or-(b2-4ac)1/2/2a

x = (43.8 +- ((43.8)2-4(14)(7.2))1/2) / 2(14) = (43.8 +- (1918.44 – 403.2)1/2) / 28 = (43.8 +- (1515.24)1/2) / 28 = (43.8 – 38.93) / 28 = 0.174 [HI] = 0.60 + 2x = 0.60 + 2(0.174) = 0.95 mol/L Ke =(0.60)2 / (0.20)(0.10) = 18 18 = (0.60 + 2x)2 / (2.1-x)(0.20-x) 18 = 0.36 + 2.4 x + 4x2 / 0.42-2.1x-0.20x+x2 7.56-41.4x+18x2 = 0.36 + 2.4 x + 4x2 14x2-43.8x+7.2=0 Use x = -b+or-(b2-4ac)1/2/2a

11. CO(g) + Cl2(g)  COCl2(g) mol/L Initial 2.00 1.00 Shift @E -x -x x 2.00 - x 1.00 - x x 2 -0.85=1.15 1-0.85=0.15 0.85 mol/L 5.00 = x / (1.00 – x)(2.00-x) 5.00 = x / 2.00 –3.00x + x2 10 – 15x + 5x2 = x 5x2 – 16 x + 10 = 0 x = (16+-(256-4(5)(10))1/2 / 10 x = 16-7.48 / 10 = 0.85

12. A(g) + B(g)  C(g) + D(g) mol/L 0.80 0.80 c -0.60 -0.60 0.60 0.60 e 0.20 0.20 0.60 0.60 (0.60) (0.60) Ke = = 9.0 (0.20) (0.20)

13. CH4(g) + 2 H2S(g)  CS2(g)+ 4 H2(g) x y z -z -2z 4z 0.100 0.030 z 4z (z) (4z)4 100 = (0.100)(0.030)2 256 z5 = (100)(0.100)(0.030)2 z = 0.129 mol/L x= 0.1 + 0.129 = 0.229 mol/L y = 0.030 + 2(0.129) = 0.288 mol/L

14. 2 NH3(g)  N2(g)+ 3 H2(g) 1.00 mol/L - 0.200 0.100 0.300 0.800 0.100 0.300 Ke = (0.300)3(.100) = 4.22 x 10-3 mol/L 0.82 b) If V increases the equilibrium shifts to side with more molecules so NH3 decreases c) Ke increases