Question

At a certain temperature, this reaction establishes an equilibrium with the given equilibrium constant, K_{c =} 3.13 x 10²⁷

3A(g) + 2B(g) $\rightleftharpoons$ 4C(g)

If, at this temperature, 1.40 mol of A and 4.00 mol of B are placed in a 1.00 L container, what are the concentrations of A, B, and C at equilibrium?

Answer 1

Answer. Intel Lencentration (A) = = 40 mol 1.00L 1.40 M (B). 4.oomol = 4.com 1.00L I la table 347. :+ 2Bq = 469 Initical 1.40 M 4.OOM since Ke= 3.13 x 1027 a very large means non goes to completion means limiting reactunt. A Consumer completely first. A consumed = t.40m B consumed - {x1.40 M = 0.933 m. B remaining = 44.00 – 0.933) M = 3.067 My c formel = x1.40 = 1,867 M achieve equilibrium caguin reaction in proceeds reverse direction and for reverse direction Kc is very very also very low low so changes in concentration is

ucal 3A + 2B o 1.4om 4.oom Initial om Assuming looy. completion OM 3.067 m 1.8 67 M Change -420 +371 +2u final 371 (3.067+24) (1.867–42) K = [c]" (A) 3 18) 3.13 x 1027 = (1-867-4X)" 3x]> (3.067+2732 for reverse reaction 40244 1.867 22 <23.067 3.13x1027 = (1.867)4 3113 (3.067)2 r = 56.867) 4 (@)* (3.067)2 (318K 1027) 3 2.48 x 10-10 Equilibrium concentrations A leq = 33 .32.48 X10-10 = 7.45 X 10M | Saleq = 745x10=10 M

I folea = 3.067 m roley = 1,867 M