SME
Physical Chemistry I: Thermodynamics & Kinetics
Chemical potential and solutions
Use this for “Mix 0.600 mol of benzene with 1.40 mol of toluene at 298.15 K. Find the Gibbs energy and the entropy of mixing.”
Chart
ΔGₘᵢₓ(0.3) ≈ −3,028
This is an example. Type your own values to start a new problem.
ΔGₘᵢₓ (J)
=
n (mol)
R
T (K)
×
(
xA
ln(
xA
)
+
xB
ln(
xB
))
ΔSₘᵢₓ
Entropy of mixing
calculated · J/K
Formulas
xB = 1 − xA
0.7 = 1 − 0.3
ΔGₘᵢₓ
=
n
×
8.314
×
T
×
(xA
×
ln(xA)
+
xB
×
ln(xB))
−3,028 = 2 × 8.314 × 298.15 × (0.3 × ln(0.3) + 0.7 × ln(0.7))
ΔSₘᵢₓ
=
−
n
×
8.314
×
(xA
×
ln(xA)
+
xB
×
ln(xB))
10.16 = −2 × 8.314 × (0.3 × ln(0.3) + 0.7 × ln(0.7))
Assumptions
•
An ideal solution: the molecules attract each other alike, so mixing gives off no heat (ΔHₘᵢₓ = 0) and only entropy drives it.
•
Both liquids start pure at the same temperature and pressure; xA + xB = 1, with n the total amount of both.
•
R = 8.314 J/(mol·K). The curve is x ln x + (1 − x) ln(1 − x) times nRT, lowest at xA = 0.5.
Step-by-step
WE KNOW
n = 2 mol, T = 298.15 K, xA = 0.3
FIND
mole fraction of B (xB), Gibbs energy of mixing (ΔGₘᵢₓ), entropy of mixing (ΔSₘᵢₓ)
Step 1 · Find mole fraction of B (xB)
Use xB = 1 − xA
The two mole fractions add to 1.
xB = 1 − 0.3
xB = 0.7
Step 2 · Find Gibbs energy of mixing (ΔGₘᵢₓ)
Use ΔGₘᵢₓ = n × 8.314 × T × (xA × ln(xA) + xB × ln(xB))
Each liquid adds its mole fraction times the ln of it; both terms are below 0. Multiply the sum by nRT.
ΔGₘᵢₓ = 2 × 8.314 × 298.15 × (0.3 × ln(0.3) + 0.7 × ln(0.7))
ΔGₘᵢₓ = 2 × 8.314 × 298.15 × (0.3 × (−1.204) + 0.7 × (−0.3567))
ΔGₘᵢₓ = 2 × 8.314 × 298.15 × ((−0.3612) + (−0.2497))
ΔGₘᵢₓ = 2 × 8.314 × 298.15 × (−0.6109)
ΔGₘᵢₓ = 16.628 × 298.15 × (−0.6109)
ΔGₘᵢₓ = 4,957.6382 × (−0.6109)
ΔGₘᵢₓ = −3,028 J → ΔGₘᵢₓ < 0 at every make-up: ideal liquids always mix on their own.
Step 3 · Find entropy of mixing (ΔSₘᵢₓ)
Use ΔSₘᵢₓ = −n × 8.314 × (xA × ln(xA) + xB × ln(xB))
The same sum times −nR. With ΔHₘᵢₓ = 0, ΔGₘᵢₓ = −TΔSₘᵢₓ.
ΔSₘᵢₓ = −2 × 8.314 × (0.3 × ln(0.3) + 0.7 × ln(0.7))
ΔSₘᵢₓ = −2 × 8.314 × (0.3 × (−1.204) + 0.7 × (−0.35667))
ΔSₘᵢₓ = −2 × 8.314 × ((−0.3612) + (−0.24967))
ΔSₘᵢₓ = −2 × 8.314 × (−0.61087)
ΔSₘᵢₓ = −16.628 × (−0.61087)
ΔSₘᵢₓ = 10.16 J/K → ΔSₘᵢₓ > 0: the mixture has more ways to arrange its molecules than the two pure liquids.
Check
Put the numbers back in. Both sides should match.
0.7 = 1 − 0.3 ✓
−3,028 = 2 × 8.314 × 298.15 × (0.3 × ln(0.3) + 0.7 × ln(0.7)) ✓
10.16 = −2 × 8.314 × (0.3 × ln(0.3) + 0.7 × ln(0.7)) ✓
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