Question

Suppose a 250. ml flask is filled with 1.8 mol of OC1, 0.70 mol of BroCl and 0.50 mol of BrCl. The following reaction becomes possible: Br2(g) + OC12(g) – Brocl(g) +BrCl(g) The equilibrium constant K for this reaction is 1.48 at the temperature of the flask. Calculate the equilibrium molarity of BroCl. Round your answer to two decimal places. x 5 ?

Answer 1

Concentration (M) = moles Volumechi volume of flask = 250 ml = 0.25L ( 1000 m 2 = 11) initial concentration om Olle : 1.8 mel = 7.2 M 0.25 L initial concentration of Broci = 2.8 M 0.70 mol = 0.252 initial concentration of Brel = 0.50 mol = 0:25 L 2.0M As que initially no amount is present in reverse direction to attain so the reaction equillibrum ICE Table concentration (MI Bra 1g) + OCl 191 o 9.2 + C tx 7.2 +X Brock (9) & Brcial 218 210 - -x 218-8 2.0-X

[Br.] = equillibrum [BroCl]- equillebrium cocle) = equillibrum Brcl) = equillibrum concentration concentration concentration concentration of Br . x M of Broc - ( 2.8-x)M el Olle - (72+2c ) M of Brel = ( 2.0->) M kce equillebaum constant - [ Brocl] [ BrCIJI [ Brz] [ 00123 => 1.48 - (2.8 -> ) (2.0-x) x 17.2 + 30 ) 0.48 x ² + 15:456 x - 5.6=0 Solving this quadratic equation, we get value of x x = - b ± √b²4ac Here, a= 0.48 E 29 ha 15, 456. > x = 0.358M = -5.6 (Brock] = (2.8 - x) M = (2,8 - 0.358) M = 2.44 M equillibrum concentration 1 molauity of BroC) = 2,44M Please Rate