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### 42-Equil-Constants-Ans

Course: CHM 2046c, Summer 2011
School: UNF
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Word Count: 757

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CONSTANT EQUILIBRIUM GCHEM II LAB CONCEPT PROBLEMS 1) Consider following equilibrium equation: H2 (g) + I2(g) 2HI(g) at 450.0 K. If 5.00E-3 moles of H2 and 2.00E-2 moles of I2 are placed in a 5.00 L flask and allowed to come to equilibrium. The flask was analyzed and equilibrium concentration of the resulting HI was found to be 1.87E-3 M. Calculate the Kc for the reaction. R eaction H2(g) + -3 I nitial...

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CONSTANT EQUILIBRIUM GCHEM II LAB CONCEPT PROBLEMS 1) Consider following equilibrium equation: H2 (g) + I2(g) 2HI(g) at 450.0 K. If 5.00E-3 moles of H2 and 2.00E-2 moles of I2 are placed in a 5.00 L flask and allowed to come to equilibrium. The flask was analyzed and equilibrium concentration of the resulting HI was found to be 1.87E-3 M. Calculate the Kc for the reaction. R eaction H2(g) + -3 I nitial (moles/L) 1.00 x 10 C hange (+/) -9.350 x 10-4 E quilibrium 6.500 x 10-5 I2(g) 2HI(g) -3 4.00 x 10 0 -9.350 x 10-4 +1.87 x 10-3 3.065 x 10-3 1.87 x 10-3 5.00 x 10-3 moles H2/5.00 L = 1.00 x 10-3 M H2 Kc = [HI]2 = (1.87 x 10-3)2 [H2] [I2] (6.50x10-5) (3.065x10-3) at 450.0 K and 2.00 x 10-2 moles I2/5.00 L = 4.00 x 10-3 M I2 = 3.4969 x 10-6 = 17.552 Kc = 17.6 1.99225 x 10-7 2) Consider the following equilibrium equation: H2(g) + Br2(g) 2HBr(g). A mixture of 1.374 grams of H2 and 70.31 grams of Br2 is heated in a 2.00 L flask to 700.0 K. At equilibrium the vessel is found to contain 0.566 grams of H2. Calculate the Kc for the reaction. R eaction I nitial (moles/L) C hange (+/) E quilibrium 1.374 grams H2 + Br2(g) .21999 .2004 .01959 2HBr(g) 0 + .4008 .4008 at 700.0 K 1 mole H2 . = .68154 moles H2/ 2.00 L = .34077 M 2.016 grams H2 1 mole Br2 . = .43998 moles Br2/ 2.00 L = .21999 M 159.8 grams Br2 1 mole H2 . = .2807 moles H2/ 2.00 L = .14037 M 2.016 grams Hr2 70.31 grams Br2 .566 grams H2 Kc = [HBr]2 [H2] [Br2] H2(g) .34077 .2004 .14037 = (.4008)2 = (.14037) (.01959) .16064 .002749 = 58.43 = Kc = 58.4 3) A mixture of H2 and N2 gas is allowed to reach equilibrium with NH3 at 450.5 K and total pressure of 9.907 atm.. The equilibrium mixture was analyzed and found to contain 7.38 atm H2 and 2.46 atm N2. Calculate the equilibrium constant. R eaction I nitial (moles/L) C hange (+/) E quilibrium N2(g) + 3H2(g) 2.46 atm 7.38 atm 2NH3 at 450.0 K 0.166 atm P total = pPH2 + pP N2 + pP NH3 9.907 = 7.38 + 2.46 + Pp NH3 [NH3]2 [N2] [H2]3 [.166]2 [2.46] [7.38]3 Kc = = Pp NH3 = 9.907 (7.38 + 2.46) = .02756 988.790 = 2.7872 x 10-5 PpNH3 = .166 atm Kc = 2.7 X 10-5 4) Consider equilibrium reaction: H2(g) + I2 (g) 2HI (g) Kc = 2.25 at 450.0 o C. A mixture of 9.00 x 10 3 moles of H2 and I2 is placed in a 3.00 L flask and allowed to come to equilibrium. What is equilibrium concentration reactants of and products ? R eaction H2 + I2 2HI at 450.0 K I nitial (moles/L) 9.00x10-3/3.00 9.00 x 10-3/3.00 0 C hange (+/) x x + 2x E quilibrium .00300 x .00300 x 2x 9.00 x 10-3 moles H2/3.00 L = .003000 M H2 and Kc = 9.00 x 10-3 moles I2/3.00 L = .003000 M I2 [HI]2 = (2x)2 = 4x2 = 2.25 = 2.00x = 1.500 [H2] [I2] (.003000 x) (.003000 x) (.00300 x)2 (.00300 x) 2x = 2(0.001286) = 0.00257 x = 0.001286 [H2] = 0.00171, [I2] = 0.00171, [HI] = 0.00257 5) Consider equilibrium equation: Br2(g) + Cl2(g) 2 BrCl(g) Kc = 1.25E-6 at 400.0 o C. If 0.350 moles Br2 in a 2.00 L flask and 0.550 moles of Cl2 in a 4.00 L flask are combined into a 8.00 L flask and allowed to reach equilibrium, what is the equilibrium concentration of the BrCl (g) ? R eaction Br2(g) + Cl2(g) 2BrCl(g) at 400.0 K I nitial (moles/L) .04375 .06075 0 C hange (+/) x x +2x E quilibrium .04375 x .06075 x 2x .350 moles Br2/8.00 L = .04375 M Br2 and .550 moles Cl2/8.00 L = .06875 M Cl2 Kc = 1.25 x10-4 = [ClBr]2 [Br2] [Cl2] 1.25 x 10-6 = 4x2 .003008 = (2x)2 note: Kc (1.25 x 10-6) < 2 powers of (.04375 x) (.06875 x) of ten compared to [Br2] .04375 x2 = 9.400 x10-10 x = .0003066 [ClBr] = 2 (.0003066) [BrCl] =6.13 x 10-4 6) Consider equation for deuterium a hydrogen isotope: D2(g) + H2(g) 2DH(g) Kc = 7.82 at 500.0 K. Initial concentration of D2 is 1.08 M and H2 is 1.05 M . What is equilibrium concentration of DH(g) ? R eaction D2(g) + H2(g) 2DH(g) at 500.0 K I nitial (moles/L) 1.08 1.05 0 C hange (+/) x x +2x E quilibrium 1.08 x 1.05 x 2x [DH]2 = [2x]2 = 4x2 = 7.82 2 [D2] [H2] [1.08 x] [1.05 x] x 2.13x + 1.134 7.82x2 16.656x + 8.867 = 4x2 3.82x2 16.656x + 8.967 = 0 (x = .6207 & ( x = 3.739 negative number) [DH] = 2(.6207) = Kc = [DH] = 1.24 7) Consider equation: 2 IBr(g) I2(g) + Br2(g) Kc = 2.80 at 250.5 o C. Suppose that 0.500 moles BrI in a 4.00 L flask reaches equilibrium . What are equilibrium concentration of IBr and Br2 ? R eaction 2 IBr(g) I2 (g) + Br2 (g) at 250.0 K I nitial (moles/L) .125 0 0 C hange (+/) 2x +x +x E quilibrium .125 2x x x 0.500 moles BrI/ 4.00 L = .125 M BrI Kc = 2.80 = [I2] [Br2] 2.80 = (x * x) 1.673 = x 1.673 (.125 2x) = x .209 3.346x = x [IBr]2 (.125 - 2x)2 (.125 - 2x) x = 0.04809 BrI = (.125 2(.04809)) = .02890 [IBr] = 0.0289 M [Br2] = .0481 M jpechunf07/2004
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