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?
(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
Mar 22, 2018
Why not just give the mols of 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")
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")