Bromine monoxide reacts according to the equilibrium 2 , NOBr(g) 2 , NO(g) + Br _ 2…
CHEMISTRY · Multi-part · A Level · Chemistry · Chemical equilibria (Le Chatelier’s principle and Kc) · Chemical equilibria (Le Chatelier’s principle and Kc)
Problem
Bromine monoxide reacts according to the equilibrium
2NOBr(g)⇌2NO(g)+Br2(g)
with
Kc=4.0×10−2
at 350K. A sealed vessel initially contains only 60 NOBr particles and 20 argon (Ar) particles. Argon is inert and does not participate in the reaction, but it is present in the diagram to represent an experimentally realistic background.
Design an equilibrium particulate diagram that satisfies all conditions below and answer the accompanying questions.
• The final diagram must contain all 80 original particles (reactive plus inert).
• The equilibrium constant must be satisfied using the smallest possible whole‐number counts for the three reactive species.
• Argon particles must remain unchanged.
Provide detailed, quantitative justification for each answer part.
(a) Write the Kc expression for the equilibrium involving the reactive species only.
(b) Beginning with 60 reactive particles (all NOBr), set up variables to represent the particle counts of NOBr, NO, and Br2 at equilibrium. Show how conservation of reactive particles and reaction stoichiometry constrain these variables.
(c) Solve for the equilibrium particle counts that satisfy Kc=4.0×10−2 while minimizing the total number of particles that must change identity. Present your algebraic work and final integers.
(d) Demonstrate numerically that the particle counts obtained in part (c) reproduce the equilibrium constant to two significant figures.
(e) Argon is ignored in the equilibrium expression, yet it appears in the diagram. Explain why inclusion of inert particles has no effect on Kc and discuss one pedagogical advantage of showing them in a particulate model.