Question

Consider the following reaction: PCl5(g)⇌PCl3(g)+Cl2(g)

Initially, 0.61 mol of PCl5 is placed in a 1.0 L flask. At equilibrium, there is 0.17 mol of PCl3 in the flask. What is the equilibrium concentration of PCl5?

Express your answer to two significant figures and include the appropriate units.

Answer 1

A balanced decomposition reaction is,

PCl₅(g)⇌PCl₃(g)+Cl₂(g)

Initially, # of moles of PCl₅(g) = 0.61 moles.

# of moles of PCl₃(g) at equilibrium = 0.17 moles.

# of moles of PCl₅(g) decomposed at equilibrium = 0.17 moles.

# of moles of PCl₅(g) left at equilibrium = 0.61-0.17 = 0.44 moles.

at equilibrium [PCl₅] = # of moles of PCl5 / Volume of container in L = 0.44 / 1.0 = 0.44 mol/L i.e. = 0.44 M

Hence,,

[PCl₅]_eqm = 0.44 M

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