For the reaction A <--> B, if the equilibrium constant is Keq = 11.7, and 4.0g of A (FW = 100g/mole) are initially placed into 5.0 L of solution. ?

(a) What are the equilibrium concentrations of A and B?
(b) How many grams of B (FW = 100g/mole) will be present at equilibrium?

1 Answer
Truong-Son N.
Mar 22, 2018

Why not just give the mols of A...?

#a)#

#("4.0 g"/"100 g/mol")/"5.0 L" = "0.008 mol/L"#

#A " "" "rightleftharpoons" "" " B#

#"I"" "0.008" "" "" "" "0#
#"C"" "-x" "" "" "+x#
#"E"" "0.008-x" "" "x#

So the mass action expression is:

#K = 11.7 = ([B])/([A])#

#= x/(0.008 - x)#

#11.7(0.008 - x) = x#

#12.7x = 0.0936#

#color(blue)([B] -= x = 7.37 xx 10^(-3) "M")#

#color(blue)([A] -= "0.008 M" - x = 6.30 xx 10^(-4) "M")#

#b)# Assuming #B# is something arbitrary with #"100 g/mol"#...

#7.37 xx 10^(-3) "mol"/cancel"L" xx 5.00 cancel"L" = "0.0369 mols B"#

#=> 0.0369 cancel"mols B" xx "100 g"/cancel"mol" = color(blue)("3.69 g B")#

Related questions
Impact of this question
329 views around the world
You can reuse this answer
Creative Commons License