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Published byErin Rodgers Modified over 3 years ago

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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.

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Example H 2 + I 2 2HI R I C E H2H2 I2I2 HI 112 0.100 0 -0.08 +0.16 0.02 0.16 Ratio, Initial, Change, Equilibrium [H 2 ] [I 2 ] Kc= [HI] 2 = [.020] [.160] 2 = 64

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PCl 3 + Cl 2 PCl 5 R I C E PCl 3 Cl 2 PCl 5 111 0.20.10 -0.08 +0.08 0.120.020.08 Ratio, Initial, Change, Equilibrium [PCl 3 ] [Cl 2 ] Kc = [PCl 5 ] = [.12] [.02] [.08] = 33.3

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HBrH2H2 Br 2 211 0.50000 +0.130 -0.260 0.2400.130 R I C E 2HBr H 2 + Br 2 [HBr] 2 Kc = [H 2 ] [Br 2 ] = [.24] 2 [.13] = 0.293

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CH 2 OH2H2 CO 111 0.10000 -0.020+0.020 0.020 0.080 R I C E CH 2 O H 2 + CO [CH 2 O] Kc = [H 2 ][CO] = [.08] [.02][.02] = 0.0050

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CO + H 2 O CO 2 + H 2 COH2OH2OCO 2 111 0.100 0 -x +x 0.10 - x x R I C E H2H2 1 0 +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 = 0.201-2.01x, 3.01x = 0.201 x=0.0668

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H 2 + I 2 2HI R I C E H2H2 I2I2 HI 112 0.200 0 -x +2x 0.2 - x 2x 2x/[0.2-x] = 7.04, 2x = 1.408-7.04x, x=0.156 [H 2 ][I 2 ] Kc = [HI] 2 = [0.2 -x] 2 [2x] 2 = 49.5 H 2 (I 2 also): 0.2 - 0.156 = 0.044 M HI: 2(0.156) = 0.312 M

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SO 3 + NO NO 2 + SO 2 SO 3 NONO 2 111 0.150 0 -x +x 0.15 - x x.707=x/[0.15-x], 0.106-0.71x=x, x=0.062 = [0.15 -x] 2 [x] 2 = 0.50 R I C E SO 2 1 0 +x x Kc = [NO 2 ][SO 2 ] [SO 3 ][NO] SO 3, NO: 0.15 - 0.062 = 0.088 M NO 2, SO 2 : = 0.062 M

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CO + H 2 O CO 2 + H 2 COH2OH2OCO 2 111 0.010 +x -x 0.01+x 0.01 - x.6325=(0.01-x)/(0.01+x), x=0.00225 = [0.01+x] 2 [0.01 - x] 2 = 0.40 R I C E H2H2 1 0.010 -x 0.01 - x Kc = [CO 2 ][H 2 ] [CO][H 2 O] CO, H 2 O: 0.010 + 0.00225 = 0.0123 M CO 2, H 2 : = 0.010 - 0.00225 = 0.0078 M

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There are essentially two types of problems which can be analyzed by ICE tables. Type 1 A. The initial or equilibrium concentration of some substances.

There are essentially two types of problems which can be analyzed by ICE tables. Type 1 A. The initial or equilibrium concentration of some substances.

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