Thermodynamics JEE 2018: Copper Oxidation and Pressure

What is the copper-oxidation question in thermodynamics JEE 2018?

The decisive step in this thermodynamics JEE 2018 numerical is to use the hydrogen-to-water pressure ratio, not hydrogen pressure alone, as the equilibrium constant. This is JEE Advanced 2018, Paper 2, Chemistry, Question 12. Its medium difficulty rating belongs to the question bank, not an official exam classification.

Copper is heated at 1250 K under nitrogen to prevent oxide formation. The nitrogen contains 1 mole % water vapour, which can oxidise copper through:

2 C u (s) + H_{2} O (g) \to C u_{2} O (s) + H_{2} (g)

Copper and copper(I) oxide are mutually immiscible. Find the natural logarithm of the minimum hydrogen partial pressure, expressed numerically in bar, needed to prevent oxidation.

Use the following data:

P_{total} = 1 b a r, R = 8 J K^{- 1} m o l^{- 1}, \ln (10) = 2.3

The supplied standard Gibbs energies at 1250 K are:

2 C u (s) + \frac{1}{2} O_{2} (g) \to C u_{2} O (s), Δ G_{1}^{\circ} = - 78000 J m o l^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g), Δ G_{2}^{\circ} = - 178000 J m o l^{- 1}

This is a numerical-answer question, not a multiple-choice question. The required entry is a natural logarithm, not the pressure itself.

How do you construct the required reaction and calculate its standard Gibbs energy?

Keep the copper-oxide formation reaction unchanged and reverse the water-formation reaction. This puts water on the reactant side and hydrogen on the product side. Reversing a reaction reverses the sign of its Gibbs energy change, so the water reaction changes from −178000 to +178000 J mol⁻¹:

H_{2} O (g) \to H_{2} (g) + \frac{1}{2} O_{2} (g), Δ G_{r e v}^{\circ} = + 178000 J m o l^{- 1}

Add it to the unchanged copper reaction. Before cancellation, the sum is:

2 C u (s) + \frac{1}{2} O_{2} (g) + H_{2} O (g) \to C u_{2} O (s) + H_{2} (g) + \frac{1}{2} O_{2} (g)

Cancel the half-mole oxygen term from both sides:

2 C u (s) + H_{2} O (g) \to C u_{2} O (s) + H_{2} (g)

Δ G^{\circ} = - 78000 + 178000 = + 100000 J m o l^{- 1}

The two supplied negative values cannot simply be added: that would combine the wrong reactions. A positive standard Gibbs energy does not rule out oxidation under every gas composition. The reaction quotient still matters.

Why does minimum hydrogen pressure correspond to equilibrium?

For forward copper oxidation, start with: $Δ G = Δ G^{\circ} + R T \ln Q$

Preventing spontaneous forward oxidation requires a non-negative Gibbs energy change. The minimum hydrogen pressure is the boundary between favourable and unfavourable forward oxidation, where the Gibbs energy change is zero:

Δ G \geq 0, Δ G_{boundary} = 0

At this boundary, the reaction quotient equals the equilibrium constant:

Q = K_{p}, 0 = Δ G^{\circ} + R T \ln K_{p}

\ln K_{p} = - \frac{Δ G^{\circ}}{R T}

Substitute the supplied values:

R T = 8 \times 1250 = 10000 J m o l^{- 1}

\ln K_{p} = - \frac{100000}{10000} = - 10

Mutually immiscible copper and copper(I) oxide form separate pure solid phases. Their activities are unity, so neither contributes a variable term to the reaction quotient.

Each gas activity is its partial pressure divided by the standard pressure. The standard-pressure factors cancel in this ratio: $p^{\circ} = 1 b a r$

Q = \frac{p_{H_{2}} / p^{\circ}}{p_{H_{2} O} / p^{\circ}} = \frac{p_{H_{2}}}{p_{H_{2} O}}

At equilibrium:

K_{p} = \frac{p_{H_{2}}}{p_{H_{2} O}}

Nitrogen has no reaction-quotient term because it does not participate in the reaction. Its presence matters through the stated gas composition and total pressure.

How does the water impurity give the answer negative 14.6?

The water partial pressure is 0.01 bar, so its logarithm contributes −4.6 when converting the equilibrium ratio into hydrogen pressure. Adding this contribution to −10 gives the required answer, −14.6.

First convert mole percentage to mole fraction, then multiply by total pressure:

y_{H_{2} O} = \frac{1}{100} = 0.01

p_{H_{2} O} = y_{H_{2} O} P_{total} = 0.01 \times 1 = 0.01 b a r

Strictly, the question’s logarithm means:

\ln (\frac{p_{H_{2}}}{1 b a r})

It is not the logarithm of a dimensional quantity. In the following shorthand, every individual pressure inside a logarithm means its numerical value in bar.

Expand the equilibrium relation:

\ln (\frac{p_{H_{2}}}{p_{H_{2} O}}) = - 10

\ln (p_{H_{2}}) - \ln (p_{H_{2} O}) = - 10

The water term is:

\ln (0.01) = \ln (10^{- 2}) = - 2 \ln (10) = - 4.6

Substituting it gives:

\ln (p_{H_{2}}) - (- 4.6) = - 10

\ln (p_{H_{2}}) = - 10 - 4.6 = - 14.6

Check the direction: at fixed water partial pressure, increasing hydrogen pressure raises the reaction quotient and Gibbs energy change, suppressing forward oxidation. Reducing hydrogen below the threshold makes the Gibbs energy change negative, so forward oxidation becomes thermodynamically favoured.

Why is negative ten an incorrect answer to this question?

Negative ten is an illustrative incorrect result, not an option from the paper. It comes from correctly calculating the equilibrium constant’s logarithm, then incorrectly treating it as the logarithm of hydrogen pressure.

The faulty step is:

\ln K_{p} = - 10 \overset{incorrect}{\to} \ln (p_{H_{2}}) = - 10

The equilibrium constant is the hydrogen-to-water partial-pressure ratio. Replacing that ratio with hydrogen pressure alone effectively assumes that water has a partial pressure of 1 bar. The supplied composition instead gives 0.01 bar.

Repair the calculation by retaining the water term, with numerical pressures in bar:

\ln (p_{H_{2}}) = \ln K_{p} + \ln (p_{H_{2} O}) = - 10 - 4.6 = - 14.6

How do you solve two related thermodynamics variations?

Use the equilibrium ratio to find a new threshold pressure, and the Gibbs energy sign to test a given gas mixture. These are original practice variations, not additional verified PYQs. Only the gas composition changes, not the reaction’s standard Gibbs energy.

Question 1: At the same temperature and total pressure, with all other data unchanged, reduce the water impurity to 0.1 mole %. Find the new minimum natural logarithm of hydrogen partial pressure, expressed numerically in bar.

The worked check is:

p_{H_{2} O} = \frac{0.1}{100} \times 1 = 0.001 b a r

\ln (p_{H_{2} O}) = - 3 \times 2.3 = - 6.9

\ln (p_{H_{2}}) = - 10 - 6.9 = - 16.9

Less water requires less hydrogen to reach the same threshold pressure ratio.

Question 2: At 1250 K, suppose the gas pressure ratio is:

\frac{p_{H_{2}}}{p_{H_{2} O}} = e^{- 9}

Calculate the Gibbs energy change for forward copper oxidation and decide whether it is thermodynamically favoured.

Δ G = 100000 + (8 \times 1250) (- 9) = + 10000 J m o l^{- 1}

Forward oxidation is not thermodynamically favoured. This sign does not establish an oxidation rate or any kinetic behaviour.

For your next attempt, write the reaction quotient first, mark the zero-Gibbs-energy boundary, and only then substitute the water partial pressure.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Keep going with Dimensional Analysis JEE 2019: Why A, B and D Are Correct.

Frequently asked questions

What is the answer to the JEE Advanced 2018 copper oxidation question?

The required numerical answer is -14.6, the natural logarithm of the minimum hydrogen partial pressure expressed numerically in bar. The equilibrium calculation gives ln Kp = -10, and the water partial pressure contributes ln(0.01) = -4.6.

Why is -10 wrong in the thermodynamics JEE 2018 question?

The value -10 is ln Kp, not the logarithm of hydrogen partial pressure. Here Kp is the hydrogen-to-water partial-pressure ratio, so the water pressure of 0.01 bar must be included. Using the supplied ln(10) = 2.3 gives -10 - 4.6 = -14.6.

Why are copper and copper oxide excluded from the equilibrium expression?

Copper and copper(I) oxide are mutually immiscible and form separate pure solid phases, each with unit activity. The reaction quotient therefore reduces to the hydrogen partial pressure divided by the water partial pressure.

Why is Gibbs energy zero at the minimum hydrogen pressure?

Preventing spontaneous forward copper oxidation requires the reaction Gibbs energy change to be non-negative. The minimum hydrogen pressure is the boundary where this change is zero and Q equals Kp. At fixed water partial pressure, adding more hydrogen raises Q and makes forward oxidation thermodynamically unfavourable.

Thermodynamics JEE 2018: Copper Oxidation Solved