Question

consider the reaction A+B<----->C+D, which has an equilibrium constant K, equal to 3.4 x 10^2

1. If one begins a reaction by placing 0.6000 moles of A in 1.0 L container as well as 0.150 moles of B, what will be the equilibrium concentrations of A, B, C, and D?

2 Once the reaction in problem one reaches equilibrium, some additional B is injected into the flask from an outside source. LeChatelier's principal says the reaction will be

a) unchanged

b) shift to left

c) shift to right

Answer 1

0 Answer A + B (+ D Volum ( - IL n = 0.6000 mole ng = 0.150 mole [A] = c = 0.6000 = 0.6000 M [B] = 0.150 - 0.750 M лв с р to 06000 0.150 (0.6000) (0.150-4) 1 2 Keq = 3.4 x10° [4167 [A] [6] 3.4 x 10² = nxn (0.6-4) (0.15-2) n²= 3.4 x 10² (0.68 0.15 - 0.154 - 0.6n + n2] n²= 3.480 6X0-15 X 202 – 0.75 X3-4 X10²n + 3.4x10² 339 2² - 255 x +30°6 = 0 ne-(-255) + (-255)2 - 4 X 339 X 30.6 2 X 339 - 255 65025 - 41,493.6 2x339 n = 255 + 23,531.9 2 x 339 = 255 + 1 53.4 2 X 33 9 = 0.14985 n = 0.6023 (not acceptables (0-15-2)=(0.15-0.6023) =-ve conce can not be negative

Equilibrium concentration = 0.6000 - 0.1499 LA leg LA Teg = 0.4501 M LB leg = 0.150 - 0.1999 = 0.0ool [Jeg = 0.1999 M [D7eg= 0.1499 M A + B (+D Acording to he' chatelic principles it we increase the concentration of B, the equilibrium tries to re-establish itself by neutralising the effect produced. Here we are increasing the concentration of B , it tries to decrease the increased concentration So the reaction shifts to the right side.