Carbonylbromide (COBr_2) is formed by the reaction between carbon monoxide and b
ID: 892952 • Letter: C
Question
Carbonylbromide (COBr_2) is formed by the reaction between carbon monoxide and bromine gas. CO(g) + Br_2(g) COBr_2(g) At equilibrium, the equilibrium constant K_p^phi = 0.18 and the partial pressures of COBr_2(g), CO(g) and Br_2(g) are 0.12,1.00, and 0.65 atm, respectively. If certain amount of bromine gas condenses to bromine liquid, and its partial pressure decreases to 0.50 atm. Which way will the equilibrium shift? Explain briefly. What are the partial pressures of all gases after equilibrium is re-established?Explanation / Answer
CO (g) + Br2 (g) <-------------------> COBr2 (g)
1.0 0.65 0.12 ----------------------> equilibrium
(a ) certain amoumt of Br2 gas condense to Br2 liquid Br2 gas partial pressure decreases on reactant side. so the reaction shifts towards backward direction.
according to Lechatliers principle if reactant partial pressure or concentration decreases reaction shifts towards backward direction if product concentration or partial pressure equilibrium shifts towards product side.
(b)
CO (g) + Br2 (g) <-------------------> COBr2 (g)
1.0 0.65 0.12 ----------------------> equilibrium
1.0- 0.5 0.65-0.5 0.12 + 0.65 --------------------> after re-establish the equilibrium
0.5 0.15 0.77
partial pressures after equilibrium re-establish
partial pressure of CO = 0.5 atm
partial pressure of Br2 = 0.15 atm
partial pressure of COBr2 = 0.77 atm