Question

Consider the reaction below. 1.48 mol of A and 8.30 mol of B are added to a 3.0 L container. At equilibrium, the concentration of A is 0.153 M. What is the value of the equilibrium constant?

3 A + 4 B ↔↔ 5 C

K_eq =

Please help!! Due tonight!!

Answer 1

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- 3 A + 4B 56 Keg = ? keq = [com Here, first we calculate the initial concentration nolarity = no. of moles of solute (mol) Volume of solution (2) 1-48mof 3.0L = 0.493 mol/L LBJ. = 8.30 mol BOL = 2.766 mol/L = 9.766 M [c] = o(because, no formation of e initially, and [AJeg = 0.153M (given data), Now, use ICE table to calculate equilibrium concentrations Initial 0.493 M 2.766 M Change -3* -48 +5x Equilibrium 0.493M.3 2.7664 X 5x. Here AJeg = 0.493-31 0.153M = 0.493M-37 → 3x = 0.493 M-0.153M = 0.34 M >> X = 0.34M = 0.1133 M Then [B]eg = 2 -766M --4x = 2.766 -400.1133 M) = 2.766M-0-453M [6] eq = 2.312 M [Cheq = 5* = 5(0-ll 33 m) = 0.566 M Now, Now, Keq = TAB 787 4 - (0.56675 00-153)3 CQ-31234 Kega 0.058 (0.00358) (28.59) 0.058 0.102