JEE Mains Chemistry · Equilibrium

Degree of Dissociation and Gibbs Energy

Expressing Kp through the degree of dissociation α and the total pressure, and linking K to ΔG° and to ΔH° through its change with temperature.

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Why this matters

Fifteen PYQs, eight of them numerical, and five from 2026. Ten link the degree of dissociation to Kp or K, usually as a formula to pick out; five turn K into ΔG° or read ΔH° from how K changes with temperature.

Concept 1 of 2: Degree of dissociation and Kp

Start with 1 mol and let a fraction α break up. Every amount is then a simple expression in α, and so is the total. Each partial pressure is its mole fraction times the total pressure P. Put these into Kp and one formula links α, P and Kp for that reaction type.

Definition

$\mathrm{A\rightleftharpoons B+C}$ : total $1+\alpha$ , $K_p=\frac{\alpha^2P}{1-\alpha^2}$ , so $\alpha=\sqrt{\frac{K_p}{K_p+P}}$ .
$\mathrm{A\rightleftharpoons 2B}$ : $K_p=\frac{4\alpha^2P}{1-\alpha^2}$ .
$\mathrm{A\rightleftharpoons B+\tfrac12C}$ : total $\frac{2+\alpha}{2}$ , $K_p=\frac{\alpha^{3/2}P^{1/2}}{(2+\alpha)^{1/2}(1-\alpha)}$ . For small α, $\alpha=\left(\frac{2K_p^2}{P}\right)^{1/3}$ .
When the products have more gas moles, a higher P lowers α.
Weak electrolyte $\mathrm{A_xB_y}$ at concentration c, α small: $K=x^xy^yc^{x+y-1}\alpha^{x+y}$ , so $\alpha=\left(\frac{K}{x^xy^yc^{x+y-1}}\right)^{1/(x+y)}$ .

A ⇌ B + C

K_p=\frac{\alpha^2P}{1-\alpha^2}\qquad\Longleftrightarrow\qquad \alpha=\sqrt{\frac{K_p}{K_p+P}}

Worked example

For

\mathrm{PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)}

K_p=1

atm. Find the degree of dissociation at a total pressure of 3 atm.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q131Moderate

Example 1 · Equilibrium · Degree of Dissociation and Gibbs Energy

A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2}C_{(g)}

. The correct relationship between

K_{P}

\alpha

and equilibrium pressure

P

Pressure pushes α down, not up

From

\alpha=\sqrt{K_p/(K_p+P)}

: when P is far larger than

K_p

, α is close to 0; when

K_p

is far larger than P, α is close to 1. The options often swap these two.

The fraction is K over the rest

For

\mathrm{A_xB_y}

\alpha=\left(\frac{K}{x^xy^yc^{x+y-1}}\right)^{1/(x+y)}

. The inverted fraction, or a power of

x+y

in place of the root, are the usual distractors.

Drill 9 more on degree of dissociation and kp

Concept 2 of 2: K, ΔG° and temperature

A reaction with a large K has a negative standard Gibbs energy change, and K = 1 means ΔG° = 0. The link is a logarithm. Temperature is the only thing that changes K, and how fast log K moves with 1/T gives ΔH°.

Definition

$\Delta G^\circ=-RT\ln K=-2.303\,RT\log K$ .
$\Delta G^\circ<0\Rightarrow K>1$ ; $\Delta G^\circ=0\Rightarrow K=1$ .
From formation values: $\Delta G^\circ=\sum\Delta_fG^\circ(\text{products})-\sum\Delta_fG^\circ(\text{reactants})$ , each times its coefficient.
Van't Hoff: $\log K=-\frac{\Delta H^\circ}{2.303R}\cdot\frac1T+\text{constant}$ . The slope of $\log K$ against $\frac1T$ is $-\frac{\Delta H^\circ}{2.303R}$ .
$\log\frac{K_2}{K_1}=\frac{\Delta H^\circ}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$ .

Gibbs energy and K

\Delta G^\circ=-2.303\,RT\log K

Worked example

A reaction has

K=100

at 300 K. Find

\Delta G^\circ

. (

R=8.314\ \mathrm{J\,K^{-1}mol^{-1}}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 2 · Q49Moderate

Example 2 · Equilibrium · Degree of Dissociation and Gibbs Energy

In a closed flask at 600 K , one mole of

X_{2}Y_{4}(\text{ }g)

attains equilibrium as given below:

X_{2}Y_{4}(\text{ }g) \rightleftharpoons 2{XY}_{2}(\text{ }g)

At equilibrium,

75\% X_{2}Y_{4}(\text{ }g)

was dissociated and the total pressure is 1 atm . The magnitude of

\Delta_{r}G^{\ominus}

(in

kJ{mol}^{- 1}

) at this temperature is

\_\_\_\_

. (Nearest Integer) (Given:

R = 8.3\text{ }J{\text{ }mol}^{- 1}{\text{ }K}^{- 1};\ln10 = 2.3

\log2 = 0.3,\log3 = 0.48,\log5 = 0.69,\log7 = 0.84)

K belongs to the equation as written

Doubling an equation squares K and doubles

\Delta G^\circ

. For HI decomposition,

\mathrm{HI\rightleftharpoons\tfrac12H_2+\tfrac12I_2}

and

\mathrm{2HI\rightleftharpoons H_2+I_2}

give answers that differ by a factor of 2. Use the equation the question writes.

Multiply the slope by 2.303

The slope of a

\log_{10}K

plot is

-\frac{\Delta H^\circ}{2.303R}

. A slope of

-100

gives

\frac{\Delta H^\circ}{R}=230.3

K, not 100.

Drill 4 more on k, δg° and temperature

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Degree of dissociation and Kp
A ⇌ B + C
$K_p=\frac{\alpha^2P}{1-\alpha^2}\qquad\Longleftrightarrow\qquad \alpha=\sqrt{\frac{K_p}{K_p+P}}$
K, ΔG° and temperature
Gibbs energy and K
$\Delta G^\circ=-2.303\,RT\log K$

Watch out for (4)

Pressure pushes α down, not up
→ Degree of dissociation and Kp
The fraction is K over the rest
→ Degree of dissociation and Kp
K belongs to the equation as written
→ K, ΔG° and temperature
Multiply the slope by 2.303
→ K, ΔG° and temperature

Test yourself on Equilibrium

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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