Chemical Equilibrium JEE 2016: Why Option C Is Incorrect

Q: What is the answer to the JEE Advanced 2016 Paper 2 chemical equilibrium question?

Option C is the incorrect statement; A, B and D are correct. The supplied information gives only the sign of the standard reaction Gibbs energy, so it does not establish the claimed exact equilibrium amount of 0.7 moles of atomic gas.

Q: Why does dissociation start spontaneously when standard Gibbs energy is positive?

Initially, no atomic product is present, so the reaction quotient is zero and lies below the finite, positive equilibrium constant. Forward dissociation is therefore spontaneous. The direction depends on the actual composition through ΔrG = ΔrG° + RT ln Qp, not on ΔrG° alone.

Q: Why does lowering pressure increase dissociation of X2?

For X2(g) ⇌ 2X(g), the product side has more gaseous moles. Lowering total pressure at fixed temperature favours this side and increases the equilibrium amount of atomic gas. The equilibrium constant remains unchanged.

Q: How do we know Kc is less than 1 in this question?

Positive standard reaction Gibbs energy implies 0 < Kp < 1. Under the stated JEE numerical convention, Δng = 1 gives Kc = Kp/(RT), with RT = 0.083 × 298 = 24.734. Therefore, Kc < 0.04043 < 1.

Q: What does beta mean in the JEE 2016 dissociation question?

Beta is the number of moles of atomic gas X formed, not the fraction of X2 dissociated. Starting with one mole of X2 and no X, the amounts are 1 − β/2 moles of X2 and β moles of X, giving 1 + β/2 total moles.

What is the chemical equilibrium question from JEE Advanced 2016 Paper 2?

C is the incorrect statement: spontaneous initial dissociation does not establish an exact equilibrium amount. This JEE Advanced 2016 chemical equilibrium question is a paragraph-based Chemistry question from JEE Advanced 2016, Paper 2, not JEE Main. The question bank rates it medium and gives an expected solving time of 90 seconds, not an official exam time limit.

An ideal gas dissociates according to: $X_{2} (g) ⇌ 2 X (g)$

The temperature, constant total pressure and sign of the standard reaction Gibbs energy are:

T = 298 K, P = 2 b a r, Δ_{r} G^{\circ} > 0

Initially, the mixture contains one mole of molecular gas and no atomic gas:

n_{X_{2}, 0} = 1 m o l, n_{X, 0} = 0

Beta denotes the number of moles of atomic gas formed, not the fraction of molecular gas dissociated: $n_{X} = β m o l$

The supplied gas constant is:

R = 0.083 L b a r K^{- 1} m o l^{- 1}

Choose the INCORRECT statement:

Why does lowering pressure make option A correct?

Lowering total pressure at fixed temperature favours the side with more gaseous moles. Here, that is the product side, so the equilibrium amount of atomic gas increases. Pressure changes the equilibrium composition, not the equilibrium constant at fixed temperature.

The official solution first checks the change in gaseous stoichiometric coefficients: $Δ n_{g} = 2 - 1 = + 1$

One mole of gaseous reactant gives two moles of gaseous product, so Le Chatelier’s principle favours the forward direction when pressure decreases. A is a correct statement and therefore is not the answer.

Why is initial dissociation spontaneous despite positive standard Gibbs energy?

The initial mixture contains no product, so its reaction quotient lies below the finite, positive equilibrium constant. Dissociation therefore starts spontaneously. Positive standard reaction Gibbs energy does not determine the direction for every mixture.

Using partial pressures, the reaction quotient is:

Q_{p} = \frac{(p_{X} / p^{\circ})^{2}}{p_{X_{2}} / p^{\circ}}

The reference pressure is the standard pressure: $p^{\circ} = 1 b a r$

Initially, the product partial pressure is zero, so the quotient is zero. For the logarithmic treatment, approach this state from a tiny positive product pressure:

p_{X} \to 0^{+} ⟹ Q_{p} \to 0^{+}

The equilibrium constant is finite and positive:

K_{p} = \exp (- \frac{Δ_{r} G^{\circ}}{R T}) > 0

Thus, initially: $Q_{p} < K_{p}$

The forward reaction is favoured until equilibrium is approached. The Gibbs energy of the actual reacting mixture explains why:

Δ_{r} G = Δ_{r} G^{\circ} + R T \ln Q_{p}

Q_{p} \to 0^{+} ⟹ R T \ln Q_{p} \to - \infty

The limiting reaction Gibbs energy is negative despite the positive standard term. This uses a limit, not a finite, defined value for the logarithm of zero. B is correct.

Why is the exact equilibrium amount in option C unjustified?

The supplied information gives only the sign of standard reaction Gibbs energy, not its numerical value or the numerical equilibrium constant. A unique equilibrium amount requires the equilibrium constant together with the starting composition and total pressure.

Following the official solution’s information-sufficiency argument, the exact claim is not justified: $β_{e q u i l i b r i u m} = 0.7$

C is the incorrect statement. Initial forward motion tells us that product forms, not its exact final amount.

How does positive standard Gibbs energy make option D correct?

The gas standard-state relation gives a pressure equilibrium constant below one. Converting with the supplied pressure and concentration conventions makes the concentration constant smaller still. Start with: $Δ_{r} G^{\circ} = - R T \ln K_{p}$

Because the gas constant and absolute temperature are positive, positive standard reaction Gibbs energy implies:

\ln K_{p} < 0 ⟹ 0 < K_{p} < 1

The pressure and concentration constants are not interchangeable. Under the usual JEE numerical convention, with pressures in bar and concentrations in moles per litre:

K_{p} = K_{c} (R T)^{Δ n_{g}}

For this reaction, substitute the gaseous mole change and supplied gas constant: $Δ n_{g} = 1$ $R T = 0.083 \times 298 = 24.734$

Therefore:

K_{c} = \frac{K_{p}}{24.734} < \frac{1}{24.734} \approx 0.04043 < 1

This textbook numerical conversion depends on the stated pressure and concentration conventions. D is correct.

A, B and D are true; C is false; the required answer is C.

What mistake leads to choosing B instead of C?

The error is substituting standard reaction Gibbs energy for reaction Gibbs energy at the actual composition. The faulty inference is:

Standard reaction Gibbs energy is positive.
Therefore, forward dissociation cannot start spontaneously.
Therefore, B must be false.

The second step ignores the composition-dependent reaction quotient. Inspect the quotient and compare it with the equilibrium constant before deciding the initial direction:

Δ_{r} G = R T \ln (\frac{Q_{p}}{K_{p}})

These statements answer different questions:

Equilibrium preference: a pressure equilibrium constant below one describes an equilibrium preference under the chosen standard states.
Current direction: a reaction quotient below the equilibrium constant establishes spontaneous forward reaction.

A mixture containing no product can move forward spontaneously even when the equilibrium constant is below one. Judge the actual mixture, not the standard Gibbs-energy sign alone.

How can you practise the three decisions separately?

Use one question each for pressure response, current reaction direction and mole accounting. These are original practice questions from the same chapter, not additional verified PYQs.

Question 1: What happens to nitrogen dioxide when total pressure decreases at fixed temperature?

The equilibrium amount of nitrogen dioxide increases. For the reaction:

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

the gaseous mole change is: $Δ n_{g} = 2 - 1 = + 1$

The product side has more gaseous moles, so lowering pressure favours it. The equilibrium constant remains unchanged at fixed temperature.

Question 2: Which direction is spontaneous when the quotient is 0.05 and the constant is 0.20?

The forward direction is spontaneous because the reaction quotient is smaller than the equilibrium constant. For the gas reaction:

K_{p} = 0.20, Q_{p} = 0.05

Substitution makes the sign explicit:

\frac{Q_{p}}{K_{p}} = \frac{0.05}{0.20} = 0.25 < 1

Δ_{r} G = R T \ln (\frac{Q_{p}}{K_{p}}) = R T \ln (0.25) < 0

The constant being below one does not prevent forward reaction from this composition. The quotient-to-constant ratio determines the current direction.

Question 3: Starting with one mole of molecular gas and no atomic gas, what are the amounts after beta moles of atomic gas form?

The molecular gas amount falls by half of beta, while the atomic gas amount becomes beta. Each mole of molecular gas dissociated produces two moles of atoms:

n_{X_{2}} = (1 - \frac{β}{2}) m o l

n_{X} = β m o l, n_{t o t a l} = (1 + \frac{β}{2}) m o l

Beta is twice the number of moles of molecular gas dissociated, not the dissociated fraction. Use Mole concept: conversions to reactions to practise this coefficient-based accounting, then write these three mole amounts before constructing the equilibrium quotient.

Frequently asked questions

What is the answer to the JEE Advanced 2016 Paper 2 chemical equilibrium question?