Question

For the following reaction, 0.500 M of COCl2 is placed in a reaction vessel. Calculate the equilibrium concentrations of all gases at equilibrium. COCl2(g) 2 CO(g) + Cl2(g) Kc = 1.70 x 10-4 4 NH3(g) + 3 O2(g) = 2 N2(g) + 6 H2O(g) When 0.256 g of NH3 and 0.480 g if O2 are initially placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 1.96 x 10 M. Calculate K, for the reaction at this temperature.

Answer 1

The given equilibriumis COCI2 (g) = CO (g) + Cl2 (g) Initial conc. of CoCl2 = 0.500 M Since no products present initially, the equilibrium will shift towards right side. Let x be the change in the concentration of each species at equilibrium ... Kc CoCl2 (8) = CO(g) + Cl2 (g) Initial 0.500 -x 0.500 - X 0 + x 0 + x change -X equilibrium [CO](C1,] – 1.70x10+ Kc = TCOCI,] xx ==1.70x10+ (0.500-x) x2 = 8.50x10--1.70x104x x² +1.70x10-47-8.50x10-* = 0 By solving the equation, we get x = 0.009135 The equilibrium Concentrations are [COCI,]=0.500 – x=0.500- 0.009135 = 0.491 M [CO]=r = 0.009135 = 9.13x10 M [Cl]]= x= 0.009135 = 9.13x10-3]M

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Number of moles of NH3 = mass/mol.wt. = 0.256 g/(17.03 g/mol) = 0.015032 mol Number of moles of O2 = mass/mol.wt. = 0.480 g/(32 g/mol) = 0.015 mol Initial concentration of NH3 = moles/volume = 0.015032 mol/1.0 L = 0.015032 M Initial concentration of O2 = moles/volume = 0.015 mol/1.0 L = 0.015 M Let x be the change in the concentration of each species at equilibrium. м 4 NH3 + 302 F . 2 N2 + 6H20 (g) Initial 0.015032 0.015 change - 4x - 3x 1 + 2x + 6x +2x 2x +6x x equilibrium 0.015032 - 4x 0.015 - 3x 6x At equilibrium, concentration. of N, = 1.96 x 10-3 M That is 2x = 1.96 x 10-3 So, x = 9.80 x 10-4 M Equilibrium concentration of NH3 = 0.015032 - 4x = 0.015032 – 4(9.80 x 10-4) = 0.011112 M Equilibrium concentration of O2 = 0.015 - 3x = 0.015 - 3(9.80 x 10-4) = 0.01206 M Equilibrium concentration of N, -2x = 1.96 x 10-3 M. Equilibrium concentration of H2O =6x = 6(9.80 x 10-4) = 5.88 x 10-3 M The equilibrium constant is given by K - []”[H,Ojó [NHz]*[02] (1.96x10-3)? (5.88x10-3) (0.011112)* (0.01206)" K=5.94x10-6 K = -= 5.94 x 10-6