1
17.5 Solving Equilibrium Problems
– Calculation of Kc(or Kp) values from measured
equilibrium concentrations (or pressures)
– Calculation of equilibrium concentrations (or
pressures) from Kc(or Kp) values
•Equilibrium tables (“ice”tables) – give the
initial, i, change of, c, and equilibrium, e,
concentrations of reactants and products
– For a general reaction: A + 2B ↔C
→[A]i, [B]i, [C]i– initial concentrations
→[A]e, [B]e, [C]e– equilibrium concentrations
→∆[A], ∆[B], ∆[C], – change in the concentrations
→[A]e= [A]i+ ∆[A] →same is valid for B and C
A + 2B
↔
C
→∆[C] = +x
→∆[A] = -∆[C]×(1 mol A/1 mol C) = -∆[C] = -x
→∆[B] = -∆[C]×(2 mol B/1 mol C) = -2∆[C] = -2x
[C]i+ x[B]i-2x[A]i-xe
+x-2x-xc
[C]i
[B]i
[A]i
i
A + 2B ↔C
[ ]
2
ii
i
2)2)([B]([A]
)([C]
[A][B]
[C]
xx
x
Kc−−
+
==⇒
i + c = e
→The equation can be used to calculate Kcif xis
known or to calculate xif Kcis known
Using Equilibrium Quantities to Calculate K
• If all equilibrium concentrations are given, substitute
in the mass action expression to find K
• If the initial concentrations and one equilibrium
concentration are given, use an ice table to find K
Example: 1.00 mol of NH3 is sealed in a 1.00 L
container and heated to 500 K. Calculate Kcfor
2NH3(g) ↔N2(g) +3H
2(g), if at equilibrium the
concentration of NH3is 0.58 M.
→[NH3]i= 1.00 mol/1.00 L = 1.00 M
→[N2]i= [H2]i= 0
→[NH3]e= 0.58 M
→[NH3]e= 1.00 –2x= 0.58
⇒x= (1.00 – 0.58)/2 = 0.21
→[N2]e= x= 0.21 M
→[H2]e= 3x= 0.63 M
3xx1.00 - 2xe
+3x+x-2xc
001.00i
2NH3(g) ↔N2(g) +3H
2(g)
[ ]
i + c = e
160
[0.58]
3][0.21][0.6
][NH
]][H[N
2
3
2
3
3
22 .Kc===⇒
