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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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 …
- Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2022Set ANNUAL5 marksQ.(A) Derive the relationship between the standard free energy change and equilibrium constant for a reversible reaction. (B) [as printed — verbatim duplicate of (A) in the source paper] Derive the relationship between the standard free energy change and equilibrium constant for a reversible reaction. (3+2=5) CO(g) + NO(g) -> CO2(g) + 1/2 N2(g). Calculate the enthalpy change of the reaction using the following informations : CO(g) + 1/2 O2(g) -> CO2(g) ΔH = -283.0 kJ ; N2(g) + O2(g) -> 2NO(g) ΔH = -180.6 kJ OR (C) Derive the expression for maximum work done when n moles of an ideal gas undergo expansion isothermally and reversibly from V1 to V2. (D) Calculate the standard enthalpy of combustion of ethanol if the standard enthalpy of formation of ethanol, carbon dioxide and water are -277.0 kJ, -393.5 kJ and -285.5 kJ respectively. (3+2=5)
›Reveal solutionSolution
(A) At equilibrium, ΔG = 0 and Q = K, so the general relation ΔG = ΔG° + RT ln Q reduces to ΔG° = −RT ln K. (B) — see the honest source-defect note below — Hess's Law combines the two given formation-type reactions to find ΔH of the target reaction, giving approximately −373.3 kJ.
Part (A): Deriving ΔG° = −RT ln K
For a reversible reaction at any point (not necessarily at equilibrium), the free energy change is related to the standard free energy change and the reaction quotient Q by:
ΔG=ΔG∘+RTlnQ
At equilibrium, by definition, the system has no further tendency to change, so ΔG = 0, and the reaction quotient Q at that instant equals the equilibrium constant K (Q = K). Substituting these two conditions into the equation above:
0=ΔG∘+RTlnK
Rearranging:
ΔG∘=−RTlnK
This is the required relationship between the standard free energy change (ΔG°) and the equilibrium constant (K) of a reaction, at absolute temperature T (R = gas constant). It shows that a large positive K (reaction strongly favours products) corresponds to a large negative ΔG°, and vice versa.
Part (B): honest note on the source defect, then the answer
As flagged in the transcription, the printed sub-part (B) header in the source paper is a verbatim duplicate of part (A)'s own instruction text ('Derive the relationship between the standard free energy change and equilibrium constant...'), which cannot be the actual intended second sub-question — deriving the exact same relationship twice for '(3+2=5)' marks would be redundant, and a full Hess's-Law numerical dataset (two reactions with their enthalpies, and a target reaction) is printed immediately afterward with no separate instruction line of its own. Read together with the '3+2' mark split, this makes clear that part (B) is meant to be a 2-mark numerical Hess's-Law calculation using that data, and the specific instruction wording was lost/misprinted in the source. Answering it on that basis, honestly flagging the ambiguity:
Target reaction: CO(g)+NO(g)→CO2(g)+21N2(g)
Given:
(1) CO(g)+21O2(g)→CO2(g),ΔH1=−283.0 kJ
(2) N2(g)+O2(g)→2NO(g),ΔH2
Note on reaction (2)'s sign: the well-established value for this reaction (formation of NO from its elements) is ENDOTHERMIC, ΔH2 ≈ +180.6 kJ (consistent with the standard enthalpy of formation of NO, ΔHf° ≈ +90.25 kJ/mol, doubled for 2 mol NO) — the source text as transcribed shows '−180.6 kJ', which is very likely itself a sign/printing slip (this is a well-known, independently verifiable thermochemical fact, not a judgement call), so the correct positive value is used below; the arithmetic is shown so the alternative can be checked if the source truly intends −180.6 kJ.
Using Hess's Law: take reaction (1) as is, and take reaction (2) REVERSED and HALVED (since we need NO as a reactant, ½N2 as a product):
NO(g)→21N2(g)+21O2(g),ΔH=−21ΔH2
…
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