Chemical Kinetics JEE 2018: Question 13 Worked Solution

What is the answer to Chemical Kinetics JEE 2018 Question 13?

The chemical kinetics JEE 2018 numerical requires the positive entry 8500, even though the standard Gibbs-energy change is negative. This is JEE Advanced 2018, Paper 2, Chemistry, Question 13, a numerical-answer question, not an MCQ.

For the reaction below, find the magnitude of the standard Gibbs-energy change at 300 K: $A (g) + B (g) ⇌ A B (g)$

The reverse activation barrier exceeds the forward barrier by:

E_{a, b} - E_{a, f} = 2 R T

The forward pre-exponential factor is four times the reverse factor. Use the supplied values:

\ln 2 = 0.7, R T = 2500 J m o l^{- 1}

The question bank rates this medium and gives an expected solving time of 120 seconds, not an official exam time limit. The required magnitude is:

| Δ G^{\circ} | = 8500 J m o l^{- 1}

How do you divide the forward and backward Arrhenius equations?

Divide the forward Arrhenius expression by the backward expression, then subtract the denominator’s entire exponent. Following the official solution’s reversible-elementary-reaction treatment, this ratio gives the equilibrium constant. Do not apply this rule automatically to arbitrary reaction mechanisms.

The subscripts identify the forward and backward directions:

k_{f}, k_{b} & : forward and backward rate constants A_{f}, A_{b} & : forward and backward pre-exponential factors E_{a, f}, E_{a, b} & : forward and backward activation energies

Write the two Arrhenius equations:

k_{f} = A_{f} \exp (- \frac{E_{a, f}}{R T})

k_{b} = A_{b} \exp (- \frac{E_{a, b}}{R T})

For the supplied treatment, divide them:

K_{e q} = \frac{k_{f}}{k_{b}} = \frac{A_{f} \exp (- E_{a, f} / R T)}{A_{b} \exp (- E_{a, b} / R T)}

Write the exponent subtraction explicitly. Subtracting the negative backward exponent makes its contribution positive:

- \frac{E_{a, f}}{R T} - (- \frac{E_{a, b}}{R T}) = \frac{E_{a, b} - E_{a, f}}{R T}

Therefore:

K_{e q} = \frac{A_{f}}{A_{b}} \exp (\frac{E_{a, b} - E_{a, f}}{R T})

Keep the order backward minus forward. It comes from dividing exponentials, not from a separate sign convention to memorise.

How do you calculate the equilibrium constant and its logarithm?

The equilibrium constant is four times the exponential of two, and its natural logarithm is 3.4 using the supplied approximation. “Backward exceeds forward” means the backward-minus-forward difference is positive, not negative:

E_{a, b} - E_{a, f} = + 2 R T, \frac{A_{f}}{A_{b}} = 4

Substitute both inputs:

K_{e q} = 4 \exp (\frac{2 R T}{R T}) = 4 e^{2}

Expand the logarithm without skipping either term:

\ln K_{e q} = \ln (4 e^{2}) = \ln 4 + \ln (e^{2}) = 2 \ln 2 + 2

Use the question’s approximation:

\ln K_{e q} = 2 (0.7) + 2 = 1.4 + 2 = 3.4

Both the prefactor ratio and the positive exponent increase the forward-to-backward rate-constant ratio. The result must therefore satisfy: $K_{e q} > 1$

A result below one would contradict these substitutions.

How do you calculate the signed Gibbs energy before its magnitude?

The signed standard Gibbs-energy change is negative; only the requested magnitude is positive. Use the Gibbs relation after calculating the logarithm, retaining its leading minus sign:

Δ G^{\circ} = - R T \ln K_{e q}

Substitute the supplied thermal-energy value and the calculated logarithm:

Δ G^{\circ} = - (2500) (3.4) = - 8500 J m o l^{- 1}

Only now apply the absolute value:

| Δ G^{\circ} | = 8500 J m o l^{- 1}

Enter 8500. Use the question’s supplied values for the thermal-energy term and the logarithm of two, rather than replacing them with more precise constants.

The sign agrees with the equilibrium constant:

K_{e q} > 1 \Rightarrow Δ G^{\circ} < 0

At equilibrium, the actual reaction Gibbs energy is zero; the standard value need not be. The composition-dependent term cancels the standard term: $Δ G = Δ G^{\circ} + R T \ln Q$

Q = K_{e q} \Rightarrow Δ G = 0

Why does reversing the activation-energy difference give the wrong answer?

Reversing the activation-energy difference changes the equilibrium constant, so taking an absolute value afterwards cannot repair the calculation. This numerical question has no supplied options. The result below is an illustrative wrong numerical answer, not an official distractor.

Suppose the exponent is incorrectly written as forward minus backward:

K_{w r o n g} = 4 \exp (\frac{E_{a, f} - E_{a, b}}{R T}) = 4 e^{- 2}

Carrying that error through gives:

\ln K_{w r o n g} = 1.4 - 2 = - 0.6

Δ G_{w r o n g}^{\circ} = - (2500) (- 0.6) = + 1500 J m o l^{- 1}

The incorrect magnitude is:

| Δ G_{w r o n g}^{\circ} | = 1500 J m o l^{- 1}

The mistake occurred while dividing exponentials. The denominator already contains a negative exponent, and that entire exponent must be subtracted. Repair the method by writing this line before simplifying:

- \frac{E_{a, f}}{R T} - (- \frac{E_{a, b}}{R T})

Check the two sign operations separately: exponent subtraction first, then the minus sign in the Gibbs relation. Take the absolute value only after both.

How do you solve three related Arrhenius–equilibrium questions?

Separate the activation-barrier and pre-exponential-factor contributions, then combine them through the logarithm. These are original practice variations, not additional verified JEE PYQs. Use the same reversible-elementary-reaction treatment for the original gaseous reaction throughout.

For every variation, take the temperature as 300 K and use:

R T = 2500 J m o l^{- 1}, \ln 2 = 0.7

Find the magnitude of the standard Gibbs-energy change in each case. Attempt each question before reading its worked answer.

What if the pre-exponential factors are equal? The backward activation barrier exceeds the forward barrier by twice the thermal-energy term:

\frac{A_{f}}{A_{b}} = 1, E_{a, b} - E_{a, f} = 2 R T

This isolates the activation-barrier contribution. Equal prefactors contribute no extra factor to the equilibrium constant:

K_{e q} = e^{2}, \ln K_{e q} = 2

Δ G^{\circ} = - (2500) (2) = - 5000 J m o l^{- 1}

| Δ G^{\circ} | = 5000 J m o l^{- 1}

What if the activation energies are equal? The forward pre-exponential factor is four times the backward factor:

E_{a, b} = E_{a, f}, \frac{A_{f}}{A_{b}} = 4

This isolates the pre-exponential-factor contribution. The exponential becomes one:

K_{e q} = 4, \ln K_{e q} = 2 (0.7) = 1.4

Δ G^{\circ} = - (2500) (1.4) = - 3500 J m o l^{- 1}

| Δ G^{\circ} | = 3500 J m o l^{- 1}

What if both contributions are smaller? Use these inputs:

\frac{A_{f}}{A_{b}} = 2, E_{a, b} - E_{a, f} = R T

This combines both contributions again. Both terms in the logarithm remain positive:

K_{e q} = 2 e, \ln K_{e q} = 0.7 + 1 = 1.7

Δ G^{\circ} = - (2500) (1.7) = - 4250 J m o l^{- 1}

| Δ G^{\circ} | = 4250 J m o l^{- 1}

If the equilibrium-to-Gibbs connection needs revision, use How to Study Equilibrium JEE: Study Order and Checks. Then redo the original question using the bank’s expected 120-second solving time as a target, writing the exponent subtraction and the signed Gibbs-energy result before entering the magnitude.

Frequently asked questions

What is the answer to JEE Advanced 2018 Paper 2 Chemistry Question 13?

The required numerical entry is 8500 because the question asks for the magnitude of the standard Gibbs-energy change in J/mol. The signed value is -8500 J/mol, calculated using the supplied values RT = 2500 J/mol and ln 2 = 0.7.

How do you find the equilibrium constant from the Arrhenius equations?

For the reversible elementary reaction treatment used in this question, divide the forward rate constant by the backward rate constant. This gives K = (A_f/A_b) exp[(E_a,b - E_a,f)/(RT)]. Substituting the prefactor ratio 4 and activation-energy difference 2RT gives K = 4e^2 and ln K = 3.4.

Why do I get 1500 instead of 8500 in this question?

The result 1500 comes from reversing the activation-energy difference in the exponent. Dividing the Arrhenius expressions gives backward minus forward, so the exponent is +2, not -2. Taking an absolute value at the end cannot correct an earlier error in the equilibrium constant.

Why is the standard Gibbs energy negative if Gibbs energy is zero at equilibrium?

At equilibrium, the actual reaction Gibbs energy is zero, but the standard Gibbs-energy change need not be zero. The relation is ΔG = ΔG° + RT ln Q; when Q = K, the composition-dependent term cancels ΔG°. Here K > 1, so ΔG° = -RT ln K is negative.