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The following concentrations were obtained for the eee of NH, from N, and ia at equilibrium at S00K, [N,]=1.5x10?M.[H.,] = 3.0 x 10° M and [NH,] = 1.2 x 102m, Calculate equilibrium constant. The equilibrium constant for the reaction, Nj... + 2H, ,<—— 2NH,,,, can be written as, = INH, __(1.2«107)"___ = 0.031810" = 3.1810? “"(N,][H,] (1.5x107)(3.0x107) At equilibrium, the concentrations of N, = 3.0 x 10°M, O, = 4.2 x 10°M and NO = 2.8 x 10°M in a sealed vessel at 800K. What will be K, forthe reaction. —_N,,. + O,,, <=> 2NO,, ; Sag. 1) ) For the reaction equilibrium constant, K, can be written as, « et. ery * [N,][0,] (3.0x107)(4.2x107) | pemattrmmhrs Manian "J 0.622 PCI,, PC/, and Ci, are at equilibrium at 500K and having concentration 1.59M PCI, 1.59M Ci, and 1.41 M PC/i,. Calculate Ke for the reaction. PC/, == PC/, + Ci, The equilibrium constant K, for the PC/, —— = PCI, + Ci, can be written as, K = [PCI,}[CZ, ] a (1.59)(1.59) “3 (PCI, ] d.4)) eegngne EnyEE 1.79 M. The value of K, = 4.24 at 800K for the reaction, Co,,, . H,0,, = =C0,, + H Calculate equilibrium concentrations of CO,, H,, CO and H,® at 800K K, it only CO and H,0 are present initially at concentrations of 0.10M each. For the reaction, co, . + H,O, 2 <== 60F i + Hy Initial concentration: 0.1M 0.1M 0 0 Let x mole per litre of each of the product be formed. a age ; (CO, ]{H,] At lib: ilibrium constant can be written as, K, = ——2——— equilibrium, equ [CO][H,0] 4.24=x?/(0.1-x)? = x? =4.24(0.01+ x? -0.2x) => x? =0.0424 +4.2x? -0.848x => 3.24x? -0.0848x +0.0424 =0 a= 3.24, b =-0.848, c = 0.0424 for quadratic equation (ax? + bx +c =0), x=(-b+ Vb? = 4ac) /2a x =0.848+ (0.848)? — 4(3.24)(0.0424) /(3.24x 2) => x = (0.848 + 0.4118) / 6.48 x, = (0.848 -- 0.4118) / 6.48 = 0.067 — x, = (0.848 + 0.4118) / 6.48 = 0.194 the value 0.194 should be neglected because it will give concentration of the reactant which is more than initial concentration. Hence the equilibrium concentrations are, [CO,]=[H,]=x = 0.067M [CO] = [HO] = 0.1 — 0.067 = 0.033M 
