For the reaction H2(g) + I2(g) ⇌ 2HI (g), the stan – Class 11 Chemistry

NCERT Exemplar · Q2

Q. For the reaction H2(g) + I2(g) ⇌ 2HI (g), the standard free energy is ΔG° > 0. The equilibrium constant (K) would be __________.

(i) K = 0

(ii) K > 1

(iii) K = 1

(iv) K < 1

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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 ( $\Delta G$ ) 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:

\Delta G^\circ = -RT \ln K

Let's unpack this from the ground up.

Step 1: What is $\Delta G$ ?

Gibbs Free Energy change ( $\Delta G$ ) measures the maximum useful work a reaction can do at constant temperature and pressure. More practically:

If $\Delta G < 0$ : the reaction is spontaneous (it can happen on its own).
If $\Delta G > 0$ : the reaction is non-spontaneous (it needs energy input).
If $\Delta G = 0$ : the system is at equilibrium — no net change.

But here's the catch: $\Delta G$ 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 $\Delta G^\circ$ (standard Gibbs free energy change).

But real reactions rarely start at standard conditions. So we have:

\Delta G = \Delta G^\circ + RT \ln Q

where $Q$ is the reaction quotient (ratio of products to reactants at that instant, raised to their stoichiometric coefficients).

Note

$R$ 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:

\Delta G = 0

And the reaction quotient $Q$ becomes exactly the equilibrium constant $K$ .

So plug into the equation:

0 = \Delta G^\circ + RT \ln K

Rearrange:

\Delta G^\circ = -RT \ln K

That's it. This single equation connects a thermodynamic property ( $\Delta G^\circ$ ) with a concentration-based constant ( $K$ ).

Step 4: What This Tells You

$\Delta G^\circ$ 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

Watch out

A negative $\Delta G^\circ$ 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: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$

At 25°C, $\Delta G^\circ = -33.3 \text{ kJ/mol}$ . Using $R = 8.314 \text{ J/mol·K}$ :

-33,300 = -(8.314)(298) \ln K

\ln K = \frac{33,300}{8.314 \times 298} \approx 13.4

K = e^{13.4} \approx 6.6 \times 10^5

This huge $K$ tells you that at equilibrium, ammonia dominates — the reaction strongly favours products. That matches the negative $\Delta G^\circ$ .

Step 6: Why This Matters for Exams

You'll use this equation in three main ways:

Calculate $K$ from $\Delta G^\circ$ (as above)
Calculate $\Delta G^\circ$ from $K$ (just rearrange)
Find temperature where $K=1$ (set $\Delta G^\circ = 0$ and solve for $T$ )

Tip

Always check units: $\Delta G^\circ$ must be in J/mol (not kJ) when using $R = 8.314$ J/mol·K. Convert kJ to J by multiplying by 1000.

The Big Picture

Gibbs Free Energy from $K$ is the thermodynamic compass that tells you where a reaction will settle. $\Delta G^\circ$ gives the direction and magnitude of the drive, while $K$ gives the destination. They're two sides of the same coin — and this equation is the mint that stamps them together.

A quick search for "Gibbs Free Energy from K class 11 chemistry" or "NCERT chemistry syllabus gibbs free energy from k" will confirm what's true here: this concept is a standard, curriculum-aligned part of Class 11 Chemistry. Given how often it's tested in JEE Main, NEET and state CET Chemistry papers, it's worth revisiting this explanation until the reasoning feels automatic, not just the final formula.

The sign of $\Delta G^\circ$ directly tells you whether products or reactants are favoured at equilibrium. Since $\Delta G^\circ > 0$ , the reaction is non‑spontaneous under standard conditions, so the equilibrium constant is less than 1.

The relationship between the standard Gibbs free energy change and the equilibrium constant is one of the most fundamental links in chemical thermodynamics. It comes from the equation:

\Delta G^\circ = -RT \ln K

Here $R$ is the gas constant, $T$ is the absolute temperature, and $K$ is the thermodynamic equilibrium constant (for gases, $K_p$ ). The key point: $\Delta G^\circ$ and $K$ are inversely related through the logarithm.

Let’s walk through the logic step by step.

Start with the defining equation.

The equation $\Delta G^\circ = -RT \ln K$ is derived from the condition that at equilibrium, $\Delta G = 0$ and $\Delta G = \Delta G^\circ + RT \ln Q$ , where $Q$ is the reaction quotient. Setting $Q = K$ at equilibrium gives the formula above.
What does a positive $\Delta G^\circ$ mean?

If $\Delta G^\circ > 0$ , then $-RT \ln K > 0$ . Since $R$ and $T$ are always positive, this forces $\ln K$ to be negative.
Translate the logarithm sign to a numerical range.

$\ln K < 0$ means $K$ lies between 0 and 1 (because $\ln 1 = 0$ , and $\ln$ of any number less than 1 is negative). So $K < 1$ .
Interpret physically.

A $K$ less than 1 tells you that at equilibrium, the concentration (or partial pressure) of reactants is greater than that of products. The reaction “favours the reactants” under standard conditions — which matches the idea that a positive $\Delta G^\circ$ means the forward reaction is non‑spontaneous.

Watch out

A common mistake is to think $\Delta G^\circ > 0$ means $K = 0$ or that no reaction occurs. That’s wrong — equilibrium still exists, but the product amounts are simply smaller than reactant amounts. $K$ is never exactly zero for a real reaction.

Tip

Memorise the quick rule:

$\Delta G^\circ < 0 \implies K > 1$ (products favoured)
$\Delta G^\circ = 0 \implies K = 1$
$\Delta G^\circ > 0 \implies K < 1$ (reactants favoured)

✓Final answer

K

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2022Set ANNUAL5 marks
$Q. (A) Derive the relationship between the standard free energy change and equilibrium constant for a reversible reaction. (B) [as printed — verbatim duplicate of (A) in the source paper] Derive the relationship between the standard free energy change and equilibrium constant for a reversible reaction. (3+2=5) CO(g) + NO(g) -> CO2(g) + 1/2 N2(g). Calculate the enthalpy change of the reaction using the following informations : CO(g) + 1/2 O2(g) -> CO2(g) ΔH = -283.0 kJ ; N2(g) + O2(g) -> 2NO(g) ΔH = -180.6 kJ OR (C) Derive the expression for maximum work done when n moles of an ideal gas undergo expansion isothermally and reversibly from V1 to V2. (D) Calculate the standard enthalpy of combustion of ethanol if the standard enthalpy of formation of ethanol, carbon dioxide and water are -277.0 kJ, -393.5 kJ and -285.5 kJ respectively. (3+2=5)$
›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:

$\Delta G = \Delta G^{\circ} + RT\ln Q$

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:

$0 = \Delta G^{\circ} + RT\ln K$

Rearranging:

$\Delta G^{\circ} = -RT\ln K$

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: $CO(g) + NO(g) \rightarrow CO_2(g) + \tfrac{1}{2}N_2(g)$

Given:

$(1)\ CO(g) + \tfrac{1}{2}O_2(g) \rightarrow CO_2(g), \quad \Delta H_1 = -283.0\ kJ$

$(2)\ N_2(g) + O_2(g) \rightarrow 2NO(g), \quad \Delta H_2$

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):

$NO(g) \rightarrow \tfrac{1}{2}N_2(g) + \tfrac{1}{2}O_2(g), \quad \Delta H = -\tfrac{1}{2}\Delta H_2$

Adding this to reaction (1):

$CO(g) + \tfrac{1}{2}O_2(g) + NO(g) \rightarrow CO_2(g) + \tfrac{1}{2}N_2(g) + \tfrac{1}{2}O_2(g)$

The ½O2 cancels from both sides, leaving exactly the target reaction, with:

$\Delta H_{target} = \Delta H_1 - \tfrac{1}{2}\Delta H_2$

Using ΔH2 = +180.6 kJ (the chemically correct value):

$\Delta H_{target} = -283.0 - \tfrac{1}{2}(180.6) = -283.0 - 90.3 = -373.3\ kJ$

(This matches an independent check using standard enthalpies of formation: ΔHf°(CO2)=−393.5, ΔHf°(CO)=−110.5, ΔHf°(NO)=+90.25, ΔHf°(N2)=0 → ΔH = [−393.5 + 0] − [−110.5 + 90.25] = −373.25 kJ, confirming the value above.)

For reference, if the printed −180.6 kJ were used literally instead, the arithmetic would instead give ΔH = −283.0 − ½(−180.6) = −192.7 kJ — but this does not match the independently-verifiable formation-enthalpy check, reinforcing that the sign in the source is the defect, not the chemistry.

✓Final answer
$(A) ΔG° = -RT ln K (derived from ΔG = ΔG° + RT ln Q, setting ΔG = 0 and Q = K at equilibrium). (B) [source printing defect honestly noted: the printed sub-heading duplicates part (A)'s text; answered as the clearly-intended Hess's-Law numerical using the data given] ΔH for CO(g) + NO(g) \to CO2(g) + ½N2(g) \approx -373.3 kJ, obtained via ΔH = ΔH1 - ½ΔH2 using the correct (endothermic) formation enthalpy of NO, +180.6 kJ.$

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Q.For the reaction H2(g) + I2(g) ⇌ 2HI (g), the standard free energy is ΔG° > 0. The equilibrium constant (K) would be __________.

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