Question

Suppose a 250 ml flask is filled with 0.30 mol of I, and 1.5 mol of HI. The following reaction becomes possible: H2(g) +12(g)=2HI(g) The equilibrium constant K for this reaction is 0.532 at the temperature of the flask. Calculate the equilibrium molarity of 12. Round your answer to two decimal places. Пм x s ?

Answer 1

Answer' Volume = 250mL = 250 x 10-3 L = 0.250L Initical Concentration. (12) = 0.30 mo! 0.250 Į = 1.2 mol = 1.2 M (HI). - - 3 mol 0.250 Į = 6omo = 6.0 mm Since initially there is no ho so present so reaction proceeds in reverse direction I CE table H₂O + I219 2HI BI Initial 1. 2M 6.0M +x Change to Equilibrium a -24 1.2+ 6-22 K- HI) 421/17 0.532 = (6-22) 2(1.2+2) 0.532 (1.2% + 32) = 36+ 472 - 24 = - 0.6384x to 0.53222 = 36 + 4322 - 24%

= 47? 0.53232 – 24% -0.63847 +36-O 7 3.4682- 24-6384 2 + 36 - 0 solving quadratis eah aut + bnto o е - bt /ь?-цас 29 a 24.6384 + 1 124.6384)2-4 (3468) (ac) 2X 3.468 u= 24.6384 + 10.37587368 . 2 X31468 na24.6384+ 10.37587368 2x 3.468 — = 5.048 My 24 27 .048 = 10.096 n=5.048 not possible because change zu can not be larger than initial concentration of HE 6.0m x= 24.6384 – 10.37587368 2x 3.468 = 2.056 m fIzlo = 1.2+3 = 1.2+ 2.057 = 3.256 M (127e4 = 326M