Q.At 60 °C, dinitrogen tetroxide is 50 per cent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Gibbs Free Energy from K
Gibbs Free Energy from K — The Bridge Between Thermodynamics and Equilibrium
Imagine you're pushing a heavy box across a rough floor. You push hard, but the box barely moves. The potential to move is there — you're applying force — but the actual motion is tiny. That's the difference between thermodynamic spontaneity (the push) and equilibrium (where the box sits, barely budging).
Gibbs Free Energy (ΔG) tells you the push — whether a reaction can happen. The equilibrium constant K tells you how far it actually goes before stopping. The equation that links them is one of the most powerful in chemistry:
ΔG∘=−RTlnK
Let's unpack this from the ground up.
Step 1: What is ΔG?
Gibbs Free Energy change (ΔG) measures the maximum useful work a reaction can do at constant temperature and pressure. More practically:
- If ΔG<0: the reaction is spontaneous (it can happen on its own).
- If ΔG>0: the reaction is non-spontaneous (it needs energy input).
- If ΔG=0: the system is at equilibrium — no net change.
But here's the catch: ΔG depends on how much reactant and product you have at any moment. It's not a fixed number.
Step 2: Standard vs. Non-standard Conditions
Chemists define a standard state (pure substances at 1 bar, 1 M concentration for solutions, 25°C usually). Under those conditions, the free energy change is called ΔG∘ (standard Gibbs free energy change).
But real reactions rarely start at standard conditions. So we have:
ΔG=ΔG∘+RTlnQ
where Q is the reaction quotient (ratio of products to reactants at that instant, raised to their stoichiometric coefficients).
R is the gas constant (8.314 J/mol·K), T is temperature in Kelvin. The ln is natural log.
Step 3: At Equilibrium — The Key Insight
At equilibrium, the reaction has no net tendency to go forward or backward. That means:
ΔG=0
And the reaction quotient Q becomes exactly the equilibrium constant K.
So plug into the equation:
0=ΔG∘+RTlnK
Rearrange:
ΔG∘=−RTlnK
That's it. This single equation connects a thermodynamic property (ΔG∘) with a concentration-based constant (K).
Step 4: What This Tells You
| ΔG∘ value | K value | Meaning |
|---|---|---|
| Negative (<0) | K>1 | Products favoured at equilibrium |
| Zero (=0) | K=1 | Equal amounts at equilibrium |
| Positive (>0) | K<1 | Reactants favoured at equilibrium |
A negative ΔG∘ does not mean the reaction is fast — only that it's thermodynamically favourable. Kinetics (activation energy) is a separate story.
Step 5: A Concrete Example
Consider the reaction: N2(g)+3H2(g)⇌2NH3(g)
At 25°C, ΔG∘=−33.3 kJ/mol. Using R=8.314 J/mol⋅K:
−33,300=−(8.314)(298)lnK …
Concept: Standard Gibbs free energy change from the equilibrium constant: ΔG∘=−RTlnKp.
Reasoning:
-
The reaction is N2O4(g)⇌2NO2(g). Let initial moles of N2O4 be 1. At 50% dissociation, moles at equilibrium: N2O4=0.5, NO2=1. Total moles =1.5.
-
Partial pressures (total pressure P=1 atm):
PN2O4=1.50.5×1=31 atm,
PNO2=1.51×1=32 atm.
-
Kp=PN2O4(PNO2)2=1/3(2/3)2=1/34/9=34. …
Find Kp from the 50% dissociation, then use ΔG⊖=−RTlnKp. With Kp=4/3 at 333 K, ΔG⊖=−796.5 J mol−1 (≈−0.80 kJ mol−1).
Set up the equilibrium
N2O4(g)⇌2NO2(g).
Start with 1 mol N2O4; degree of dissociation α=0.5:
- N2O4=1−0.5=0.5 mol
- NO2=2×0.5=1.0 mol
- total =1.5 mol
Partial pressures (total pressure =1 atm)
pN2O4=1.50.5×1=31 atm,pNO2=1.51.0×1=32 atm.
Equilibrium constant
Kp=pN2O4pNO22=1/3(2/3)2=1/34/9=34≈1.333.
Standard free energy change …
- COMEDK 2024Set 2024-M1 markMCQQ.Given: ΔG0f of C2H2 is 2.09×105 J/mol and ΔGf0 of C6H6 is 1.24×105 J/mol. Calculate the equilibrium constant for the cyclic polymerisation of Ethyne to Benzene at 27∘C. (R=8.314JK−1 mol−1) (A) 1.1134×1010 (B) 0.1113 (C) 1.429×1088 (D) 88.15
›Reveal solutionSolution
For 3C2H2→C6H6, ΔG∘=−5.03×105 J/mol, giving lnK≈202, i.e. K∼1088 — the only option of that magnitude is (C) 1.429×1088.
Step 1 — Reaction and ΔG∘
Cyclic trimerisation of ethyne to benzene:
3C2H2→C6H6
ΔGrxn∘=ΔGf∘(C6H6)−3ΔGf∘(C2H2)
=1.24×105−3(2.09×105)=1.24×105−6.27×105=−5.03×105 J/mol
Step 2 — Relate to the equilibrium constant
ΔG∘=−RTlnK⇒lnK=RT−ΔG∘
At T=27∘C=300 K:
lnK=8.314×3005.03×105=2494.25.03×105≈201.7
Step 3 — Evaluate K …
- KCET 2022Set B-31 markMCQQ.The volume of 2.8g of CO at 270C and 0.821 atm, pressure is (R - 0.08210 lit.atm.K−1mol−1) (A) 3 litres (B) 30 litres (C) 0.3 litres (D) 1.5 litres
›Reveal solutionSolution
Convert the mass of CO to moles, convert 27∘C to kelvin, then apply PV=nRT.
1. Moles of CO
Molar mass of carbon monoxide:
M(CO)=12+16=28 g mol−1
n=Mw=28 g mol−12.8 g=0.1 mol
2. Temperature in kelvin
The gas laws are only valid on the absolute scale — a Celsius value would make T meaningless (it can even be zero or negative):
T=27+273=300 K
3. Apply the ideal gas equation
PV=nRT⟹V=PnRT
The value of R given is 0.0821 L atm K−1mol−1, which pairs with pressure in atm and volume in litres — and the pressure is already in atm (0.821 atm), so no conversion is needed.
V=0.821 atm(0.1 mol)(0.0821 L atm K−1mol−1)(300 K)
4. Arithmetic …
- KCET 2022Set B-31 markMCQQ.The work done when 2 moles of an ideal gas expands reversibly and isothermally from a volume of 1L to 10L at 300K is (R - 0.0083 kJ K mol−1) (A) 0.115 kJ (B) 58.5 kJ (C) 11.5 kJ (D) 5.8 kJ
›Reveal solutionSolution
Apply the reversible-isothermal work formula w=−2.303nRTlog10(V2/V1) — the ten-fold expansion makes the log exactly 1.
1. Why this formula (and not w=−PΔV)
In a reversible expansion the external pressure is kept infinitesimally below the gas pressure at every instant, so Pext=Pgas=nRT/V throughout — P is not constant, and we must integrate:
w=−∫V1V2PdV=−∫V1V2VnRTdV
Because the process is isothermal, T is a constant and comes out of the integral:
w=−nRTlnV1V2=−2.303nRTlog10V1V2
(The minus sign is the IUPAC convention: work done by the expanding gas is negative for the system.)
2. Insert the data
- n=2 mol
- R=0.0083 kJ K−1mol−1 (given in kJ, so the answer lands directly in kJ — no ÷1000 needed)
- T=300 K
- V1=1 L,V2=10 L⇒V1V2=10
log10(10)=1
3. Arithmetic
∣w∣=2.303×2×0.0083×300×1
Step by step:
2×0.0083=0.0166
0.0166×300=4.98
2.303×4.98=11.47 kJ …
- COMEDK 2021Set 20211 markMCQQ.The value of ΔG∘ for the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. The value of KC at 298 K is (A) 7.72×10−4 (B) 5.62×10−4 (C) 4.81×10−3 (D) 3.81×10−3
›Reveal solutionSolution
So Kc = 3.81 x 10^-3 - the small K reflects that phosphorylation of glucose is non-spontaneous on its own (in the cell it is coupled to ATP hydrolysis).
Concept: DeltaG(standard) = -RT ln K.
DeltaG = +13.8 kJ/mol = 13800 J/mol, T = 298 K, R = 8.314 J K^-1 mol^-1.
ln K = -DeltaG / (RT) = -13800 / (8.314 x 298)
RT = 8.314 x 298 = 2477.6 J/mol
ln K = -13800 / 2477.6 = -5.5698
K = e^(-5.5698)
e^(-5.5698) = e^(-5) x e^(-0.5698) = (6.738 x 10^-3)(0.5656) = 3.81 x 10^-3 …
- KCET 2018Set A-11 markMCQQ.Edge length of a cube is 300 pm. Its body diagonal would be (A) 600 pm (B) 423 pm (C) 519.6 pm (D) 450.5 pm
›Reveal solutionSolution
Apply Pythagoras twice: face diagonal =2a, body diagonal =3a, so 3×300 pm=519.6 pm.
Step 1 — Why 3a?
Take a cube of edge a with one corner at the origin. The opposite corner is at (a,a,a). The distance between them is
d=a2+a2+a2=3a
Equivalently: the face diagonal is a2+a2=2a, and the body diagonal is the hypotenuse of a right triangle whose legs are the face diagonal and one edge: (2a)2+a2=3a.
Step 2 — Substitute a=300 pm.
d=3×300=1.7320×300=519.6 pm …
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