Here are some additional problems. Some of these we have worked in class. If you play the slideshow, the given information will be shown, then you can complete the problem and click through to the answer. FYI: The R in the RICE table is just the coefficient of the compound from the reaction.

Example H 2 + I 2 2HI R I C E H2H2 I2I2 HI Ratio, Initial, Change, Equilibrium [H 2 ] [I 2 ] Kc= [HI] 2 = [.020] [.160] 2 = 64

PCl 3 + Cl 2 PCl 5 R I C E PCl 3 Cl 2 PCl Ratio, Initial, Change, Equilibrium [PCl 3 ] [Cl 2 ] Kc = [PCl 5 ] = [.12] [.02] [.08] = 33.3

HBrH2H2 Br R I C E 2HBr H 2 + Br 2 [HBr] 2 Kc = [H 2 ] [Br 2 ] = [.24] 2 [.13] = 0.293

CH 2 OH2H2 CO R I C E CH 2 O H 2 + CO [CH 2 O] Kc = [H 2 ][CO] = [.08] [.02][.02] =

CO + H 2 O CO 2 + H 2 COH2OH2OCO x +x x x R I C E H2H x x = 4.06 = [0.10 -x] 2 [x] 2 Kc = [CO 2 ][H 2 ] [CO][H 2 O] x/[0.10-x] = 2.01, x = x, 3.01x = x=0.0668

H 2 + I 2 2HI R I C E H2H2 I2I2 HI x +2x x 2x 2x/[0.2-x] = 7.04, 2x = x, x=0.156 [H 2 ][I 2 ] Kc = [HI] 2 = [0.2 -x] 2 [2x] 2 = 49.5 H 2 (I 2 also): = M HI: 2(0.156) = M

SO 3 + NO NO 2 + SO 2 SO 3 NONO x +x x x.707=x/[0.15-x], x=x, x=0.062 = [0.15 -x] 2 [x] 2 = 0.50 R I C E SO x x Kc = [NO 2 ][SO 2 ] [SO 3 ][NO] SO 3, NO: = M NO 2, SO 2 : = M

CO + H 2 O CO 2 + H 2 COH2OH2OCO x -x 0.01+x x.6325=(0.01-x)/(0.01+x), x= = [0.01+x] 2 [ x] 2 = 0.40 R I C E H2H x x Kc = [CO 2 ][H 2 ] [CO][H 2 O] CO, H 2 O: = M CO 2, H 2 : = = M