SME
Grade 10 Science · Chemistry
Reaction rates and chemical equilibrium
Use this for “N₂O₄ starts at 1 mol/L, and 0.2 mol/L of it turns into NO₂. What is K at equilibrium?” with an ICE table.
Diagram
At equilibrium Q = K = 0.2: [N₂O₄] = 0.8 M, [NO₂] = 0.4 M.
N₂O₄ that reacted: 0.2 mol/L
This is an example. Type your own values to start a new problem.
K
=
[NO₂]
2
[N₂O₄]
[N₂O₄]₀
N₂O₄ at the start
calculated · mol/L
x
N₂O₄ that reacted
calculated · mol/L
Formulas
x
=
[NO₂]/2
0.2 = 0.4/2
[N₂O₄]
=
[N₂O₄]₀
−
x
0.8 = 1 − 0.2
K
=
[NO₂]
2
/[N₂O₄]
0.2
=
0.4
2
/0.8
Assumptions
•
At equilibrium the forward and reverse rates are equal; amounts stop changing, reactions don’t stop.
•
K depends only on temperature. Pure solids and liquids are left out of K.
•
The flask starts with N₂O₄ only; x is the N₂O₄ that reacts.
Step-by-step
WE KNOW
[N₂O₄] = 0.8 mol/L, [NO₂] = 0.4 mol/L
FIND
N₂O₄ that reacted (x), N₂O₄ at the start ([N₂O₄]₀), equilibrium constant (K)
Step 1 · Find N₂O₄ that reacted (x)
Use x = [NO₂]/2
Each
N₂O₄
that
reacts
makes
two
NO₂,
so
x
is
half
the
NO₂.
x
=
0.4/2
x
=
0.2
mol/L
Step 2 · Find N₂O₄ at the start ([N₂O₄]₀)
Use [N₂O₄] = [N₂O₄]₀ − x
Add back what reacted.
[N₂O₄]₀
=
[N₂O₄]
+
x
[N₂O₄]₀ = 0.8 + 0.2
[N₂O₄]₀ = 1 mol/L
Step 3 · Find equilibrium constant (K)
Use K = [NO₂]²/[N₂O₄]
Products over reactants at equilibrium, each to the power of its coefficient.
K
=
0.4
2
/0.8
K
=
0.16/0.8
K
=
0.2
Check
Put the numbers back in. Both sides should match.
0.2 = 0.4/2 ✓
0.8 = 1 − 0.2 ✓
0.2
=
0.4
2
/0.8
✓
Topic check
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