5 RE pg CH2O H2 + CORICECH2OH2CO1110.100-0.020+0.020+0.0200.0800.0200.020[CH2O]Kc =[H2][CO]=[.08][.02][.02]=
6 14.9 - pg. 570 CO + H2O CO2 + H2 R I C E CO H2O CO2 H2 1 1 1 1 0.100 Read Follow sample calculation carefully.pg CO + H2O CO2 + H2RICECOH2OCO2H211110.1000.100-x-x+x+xxxxxKc =[CO2][H2][CO][H2O]=[0.10 -x]2[x]2= 4.06x/[0.10-x] = 2.01, x = x, 3.01x = 0.201x=0.0668PE 11, 14.40, pg. 589 (notice [ ] for 14.41)
7 PE 11 - pg. 571 H2 + I2 2HI R I C E H2 I2 HI 1 1 2 0.200 0.200 -x -x -x-x+2x0.2 - x0.2 - x2x[H2][I2]Kc =[HI]2=[0.2 -x]2[2x]2= 49.52x/[0.2-x] = 7.04, 2x = x, x=0.156H2 (I2 also): = MHI: 2(0.156) = M
8 14.40 - SO3 + NO NO2 + SO2 R I C E SO3 NO NO2 SO2 1 1 1 1 0.150 0.150-x-x+x+xxxxxKc =[NO2][SO2][SO3][NO]=[0.15 -x]2[x]2= 0.50.707=x/[0.15-x], x=x, x=0.062SO3, NO: = MNO2, SO2: = M
9 14.41 - CO + H2O CO2 + H2 R I C E CO H2O CO2 H2 1 1 1 1 0.010 0.010 +x+x-x-x0.01+x0.01+xxxKc =[CO2][H2][CO][H2O]=[0.01+x]2[ x]2= 0.40.6325=(0.01-x)/(0.01+x), x=CO, H2O: = MCO2, H2: = = M
10 Equilibrium calculations when Kc is very small Thus far, problems have been designed so that the solution for x is straightforwardIf the problems were not so carefully designed we might have to use quadratic equation (or calculus) to solve the problem.If Kc is very large or very small we can use a simplification to make calculating x simpleSetting up the RICE chart is the same, but the calculation of Kc is now slightly differentRead pg. 572, 573
11 Equilibrium calculations when Kc is small Looking at the equilibrium law for 14.10:4x3= small Kc[ x]2For Kc to be small, top must be small, bottom must be large (relative to top)For top to be small, x must be smallIf x is small, then x 0.100Notice that we can only ignore x when it is in a term that is added or subtracted.Can we ignore x in: 4x, 3+x, 0.1-3x, 3x-x, x2+1?We can for these:Try PE 12 (573). Concentrations are [initial].
12 PE 12 - pg. 573 N2 + O2 2NO R I C E N2 O2 NO 1 1 2 0.033 0.00810 -x -x-x+2x0.033-xx2x[N2][O2]Kc =[NO]2=[0.033-x][ x][2x]2= 4.8 x 10-31
13 This is the equilibrium [NO2] PE 12 - pg N2 + O2 2NO[0.033-x][ x][2x]2= 4.8 x 10-31Small Kc: Thus numerator is small and x must be small: x is negligible when adding or subtracting[0.033][0.0081][2x]2= 4.8 x 10-31[2x]2= 1.28 x 10-342x= 1.13 x 10-17This is the equilibrium [NO2]
14 R I C E HCl H2 Cl2 2 1 1 2 1 -2x +x +x 2-2x 1+x x 2HCl H2 + Cl2 Kc= 3.2 x 10–34 determine [equil], if [initial] are 2.0 M, 1.0 M, 0 MRICEHClH2Cl221121-2x+x+x2-2x1+xx[HCl]2Kc =[H2] [Cl2]=[2-2x]2[1+x] [x]=2 [x]= 3.2 x 10–34x = (3.2 x 10–34)(4) = 1.3 x 10–33[equil] are 2, 1 and 1.3 x 10–33
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