CHAPTER 14. CHEMICAL EQUILIBRIUM
Section 14.1 The Concept of Equilibrium
Chemical reactions do NOT go to completion (100% products) even those that look like
they do. A reaction will instead reach a point after which the amount of reactants and
products no longer changes with time. This is due to the fact that all reactions are reversible.
Example N O (g) ⇌ 2 NO (g)
2 4 2
(⇌ means the reaction goes in both directions)
Add some colorless N O 2g)4to an empty flask and watch it with time the gas will slowly
go more and more brown as N O (g)2co4verts to brown NO (g). 2
After a few minutes, no further color change occurs – “the reaction has reached
equilibrium”. If we analyze the gas, we find it contains both N O and2NO4. Why does2’t it
keep giving more NO and going more brown?? Because the rate at which N O is giving more
2 2 4
NO 2s now the same as the rate of the reverse reaction (2 NO combin2ng to give N O ). Thus,2 4
the [N 2 ]4and [NO ] c2nc’s do not change. We can
see this in a conc ([ ]) vs time (t) plot.
At any given time t:
Rate (forward) = k [Nfwd] 2 4
= Δ[N2O 4/Δt
Rate (reverse) = k rev ] 2
= Δ[NO 2/Δt
(assume reaction is elementary)
The plot shows that after a certain time, the
conc’s of all species do not change any more. The
reaction has reached equilibrium. This is a “dynamic
equilibrium”, which means reactions are still
occurring in both directions, but no overall change
in the [ ]’s.
Sections 14.2 and 14.3. Equilibrium Constant (K) and Reaction Quotient
At equilibrium: rate (forward) = rate (reverse)
∴ k fwdO 2 =4k [NOrev 2
∴ k /k = [NO ] /[N O ] = K
fwd rev 2 2 4
∴ K = k /fwd rev
1 K is the Equilibrium Constant for the reaction
(do NOT confuse equilibrium constant K (capital letter K) with rate constants k and k fwd rev
(small letter k).
Similarly: H 2g) + I 2(g) ⇌ 2 HI (g)
at equilibrium, k [H ][I ] = k [HI] 2
fwd 2 2 rev 2
∴ k fwd =revI] /[H ][I ] =2K 2
The equilibrium constant (K) is a number (no units!) that is determines the relative amounts of
products and reactants at equilibrium:
Note (i) **Only temperature, T, can change the value of K (see later)**
(ii) “reactants” are defined as the things on the left, and “products” as the things on the
right of the balanced equation.
Now, consider again the reaction: H (g) + I (g) ⇌ 2 HI (g)
At equilibrium: K = [HI] /[H ][2 ] 2
But, if the reaction is not yet at equilibrium: [HI] /[H ][I ] =2Q (2 K)
Q is the “reaction quotient” or “massaction expression”
As the reaction proceeds towards equilibrium, Q is changing. When the reaction reaches
equilibrium, Q = K.
Remember!! At equilibrium, Q = K
Writing Q and K for any reaction
Consider a A + b B ⇌ c C + d D
Q c= [C] [D] d a, b, c, d are stoichiometry
[A] [B] coefficients in the balanced eqn.
(Q cmeans Q expressed in [ ]’s).
At equilibrium, Q = c c
So, all we have to do is:
1) Write the equation
2 2) Balance it !!
3) Write down Q and K
Example 1: N (g) +23H (g) 2 ⇌ 2NH 3g) check: it is balanced
∴ Q =c[NH ] /[3 ][H ]2 (=2K if atcequilibrium)
C 3 8g) + O 2g) ⇌ CO 2(g) + H O2(g) check: not balanced, so we must balance it first!
C H (g) + 5O (g) ⇌ 3CO (g) + 4H O (g)
3 8 2 2 2
∴ Q =c[CO ] [2 O] 2[C H ] [3 ]8 (=2K if at ecuilibrium)
i) Q and K are just numbers (i.e. no units) because each conc is divided by standard conc =
1.00 M, so the units (M) cancel.
ii) Q and K refer to a balanced reaction as written (important! see below)
iii)No matter how many steps the reaction takes to get to products, we need only the
overall balanced equation to get Q and K
Section 14.4. The Meaning of the Equilibrium Constant (K)
Consider again: N O (g) ⇌ 2 NO (g) at equilibrium
2 4 2
K = [NO ] 2
[N 2 ] 4
If K very large: [NO ] >> [2 O ] 2e4say the equilibrium lies far to
the right i.e. mainly products (things on the right of equation)
If K very small: [NO ] <> 10 products favored (“equilibrium lies to the right”)
K < K reaction proceeds from righttoleft to reach equil