SME
Grade 10 Science · Problem type
Le Châtelier: Q against K
Use this for “H₂ is added to H₂ + I₂ ⇌ 2HI at equilibrium. Which way does it shift?”
Diagram
At equilibrium Q = K = 49: [H₂] = 0.1 M, [I₂] = 0.1 M, [HI] = 0.7 M.
After the change Q = 24.5 is less than K = 49: the reaction shifts right, making more products until Q = K again.
This is an example. Type your own values to start a new problem.
[H₂]
H₂ at equilibrium
example · mol/L
[I₂]
I₂ at equilibrium
example · mol/L
[HI]
HI at equilibrium
example · mol/L
K
Equilibrium constant
calculated
a
H₂ added
example · mol/L
Q
Reaction quotient just after
calculated
Formulas
K
=
[HI]
2
/([H₂]
×
[I₂])
49
=
0.7
2
/(0.1
×
0.1)
Q
=
[HI]
2
/(([H₂]
+
a
)
×
[I₂])
24.5
=
0.7
2
/((0.1
+
0.1)
×
0.1)
Assumptions
•
The reaction runs in a closed container at one temperature, so K stays the same.
•
Concentrations are in mol/L; K has no unit here.
•
Q is K’s formula with the concentrations right after the change: Q below K shifts toward products.
Step-by-step
WE KNOW
[H₂] = 0.1 mol/L, [I₂] = 0.1 mol/L, [HI] = 0.7 mol/L, a = 0.1 mol/L
FIND
equilibrium constant (K), reaction quotient just after (Q)
Step 1 · Find equilibrium constant (K)
Use K = [HI]²/([H₂] × [I₂])
Products over reactants at equilibrium.
K
=
0.7
2
/(0.1
×
0.1)
K
=
0.7
2
/0.01
K
=
0.49/0.01
K
=
49
Step 2 · Find reaction quotient just after (Q)
Use Q = [HI]²/(([H₂] + a) × [I₂])
Just
after
the
H₂
goes
in,
only
[H₂]
has
changed.
Compare
Q
with
K
.
Q
=
0.7
2
/((0.1
+
0.1)
×
0.1)
Q
=
24.5
is
below
K
=
49,
so
the
reaction
shifts
toward
HI
(the
product)
Q
=
24.5
Check
Put the numbers back in. Both sides should match.
49
=
0.7
2
/(0.1
×
0.1)
✓
24.5
=
0.7
2
/((0.1
+
0.1)
×
0.1)
✓
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