Question

Suppose a 500 ml flask is filled with 2.0 mol of CHCI, 0.50 mol of HCl and 1.5 mol of CCI. The following reaction becomes prie CI (8) + CHCI (8) -HCI(g)+CCI (8) The equilibrium constant for this reaction is 0.352 at the temperature of the flask Calculate the equilibrium molarity of CHCI, Round your answer to two decimal places

Answer 1

No of molesol solute Answer! We know, molarity (M) - volume of the solution in litre 2 CH C13] Initiae = som xnó}) (:1mL = 1634) . = 4 mol L = 4M (.:Mmulle) .: [HU] Initial = _0.50 mol = 1 mollectin (500x103) L . [cad]riline = los mod - 3 mol/l = 3M . . (soox 103) 2 The given eduation is as follows: 2 (g) + CHCl3 (9) Ź Hel (9) + cca (9) k=0.352 The equilibrium constant (k) can be expressed in terms of concentration as k= [Ha ] [ccla ] [42] [CHCl37 The reverse readion is as follows - НИ (9) +cl4t9) Л, 14) + сниз (9) .: The equilibrium constant k'. [42] [CH3 - F inna [HQ] [204] K = 284 Now we have to construct ICE table which is as follows: R HU I ela I a I CHCl3 Il 1M 3M T OM I 4M cl ann anm T tum +xM El (1-x) I (3-2) n I nm 4+x) x4 .. k' [42] [CH Ulz] [HU] [ccla] or, 2.84 = m (4 tu) antn? (1-1) (3-1) 3-34-x tu2 n² 4x43 a, 4h +M2 = 2184 (2ax +3) = 21841?_ 11,368 + 8.52 ar, MF_2.84? + ax + 11636X - 8152 = 0 a, -1184 1? + 15:36% -8.52 = 0 ar 1184 x² - 15:36h 48.52=0 this is a quadratic equation. For a general duadratic equation, an²tbute=0 0.352 - = 4x+n2

x = -b : Vipe tac ac si laitteeseen ere, a = 1.84, 84, b= - 15:36, C-8152: b= - 15:36, C-852 sy= =(-15:36) + VC-15:36)2-4*184x852 2x1.84 in = 15:36 + 13.16 .. , My = 15:36-13-16 3.68 = 7.75 The bolt 368 is not accepted as it is higher ^2 = 0.597 À This roue is releptet than 3. .[CHC3] = (4+x) M = (4+0.597) M = 4.597 M " The equilibrium molarity of cHUz is 4.6 M. (Andre)