SME
General Chemistry II
Equilibrium
Use this for “K = 50.5 for H₂ + I₂ ⇌ 2HI. A flask starts with 0.100 M H₂, 0.100 M I₂ and 0.0500 M HI. Find every concentration at equilibrium.”
Diagram
At equilibrium Q = K = 50.5: [H₂] = 0.0275 M, [I₂] = 0.0275 M, [HI] = 0.195 M.
This is an example. Type your own values to start a new problem.
K
Equilibrium constant
example
[H₂]₀
Starting H₂
example · M
[I₂]₀
Starting I₂
example · M
[HI]₀
Starting HI
example · M
x
Change (forward)
calculated · M
[H₂]
H₂ at equilibrium
calculated · M
[I₂]
I₂ at equilibrium
calculated · M
[HI]
HI at equilibrium
calculated · M
Formulas
K
=
[HI]²
÷
([H₂]
×
[I₂])
50.5
=
0.195
2
0.0275 × 0.0275
[H₂]
=
[H₂]₀
−
x
0.0275 = 0.1 − 0.0725
[I₂]
=
[I₂]₀
−
x
0.0275 = 0.1 − 0.0725
[HI]
=
[HI]₀
+
2
×
x
0.195 = 0.05 + 2 × 0.0725
([HI]₀ + 2 ×
x
)
2
=
K
×
([H₂]₀
−
x
)
×
([I₂]₀
−
x
)
(0.05 + 2 × 0.0725)
2
=
50.5
×
(0.1
−
0.0725)
×
(0.1
−
0.0725)
Assumptions
•
Only the concentrations at equilibrium go into K, never the starting ones.
•
x is how far the reaction runs forward: each step uses one H₂ and one I₂ and makes two HI. A negative x runs it backward.
•
K = [HI]² ÷ ([H₂][I₂]) is a quadratic in x. Of its two roots, keep the one that leaves no concentration below 0.
•
One temperature throughout, so K stays the same. Concentrations are in mol/L (M).
Step-by-step
WE KNOW
K = 50.5, [H₂]₀ = 0.1 M, [I₂]₀ = 0.1 M, [HI]₀ = 0.05 M
FIND
change (forward) (x), H₂ at equilibrium ([H₂]), I₂ at equilibrium ([I₂]), HI at equilibrium ([HI])
Step 1 · Find change (forward): x
Use ([HI]₀ + 2 × x)² = K × ([H₂]₀ − x) × ([I₂]₀ − x)
Put
the
equilibrium
amounts
into
K
:
([HI]₀ + 2
x
)
2
=
K
([H₂]₀
−
x
)([I₂]₀
−
x
).
Expand
it
into
ax²
+
bx
+
c
=
0
and
use
the
quadratic
formula.
x
=
(−10.3
+
√(
10.3
2
−
4
×
(−46.5)
×
(−0.5025)))
÷
(2
×
(−46.5))
The
roots
of
−
46.5
x
²
+
10.3
x
−
0.5025
=
0,
in
order:
0.07255,
0.149.
Rejected:
0.149,
since
x
must
leave
every
concentration
at
least
0.
x
=
+0.0725
M
Step 2 · Find H₂ at equilibrium ([H₂])
Use [H₂] = [H₂]₀ − x
One H₂ is used for each step of the reaction.
[H₂] = 0.1 − 0.0725
[H₂] = 0.0275 M
Step 3 · Find I₂ at equilibrium ([I₂])
Use [I₂] = [I₂]₀ − x
One I₂ is used with each H₂.
[I₂] = 0.1 − 0.0725
[I₂] = 0.0275 M
Step 4 · Find HI at equilibrium ([HI])
Use [HI] = [HI]₀ + 2 × x
Two HI form for each H₂ used.
[HI] = 0.05 + 2 × 0.0725
[HI] = 0.05 + 0.145
[HI] = 0.195 M
Check
Put the numbers back in. Both sides should match.
50.5
=
0.1951
2
0.02745 × 0.02745
✓
0.0275 = 0.1 − 0.0725 ✓
0.0275 = 0.1 − 0.0725 ✓
0.195 = 0.05 + 2 × 0.0725 ✓
(0.05 + 2 × 0.07255)
2
=
50.5
×
(0.1
−
0.07255)
×
(0.1
−
0.07255)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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