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JEE Advanced 2018 · Paper 2, Chemistry question 12

The surface of copper gets tarnished by the formation of copper oxide. N2\mathrm{N_2} gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the N2\mathrm{N_2} gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:

2Cu(s)+H2O(g)→Cu2O(s)+H2(g)\mathrm{2Cu(s) + H_2O(g) \rightarrow Cu_2O(s) + H_2(g)}

pH2p_{\mathrm{H_2}} is the minimum partial pressure of H2\mathrm{H_2} (in bar) needed to prevent the oxidation at 1250 K. The value of ln⁡(pH2)\ln(p_{\mathrm{H_2}}) is ____.

(Given: total pressure = 1 bar, RR (universal gas constant) = 8 J K−1^{-1} mol−1^{-1}, ln(10) = 2.3. Cu(s) and Cu2_2O(s) are mutually immiscible.

At 1250 K: 2Cu(s)+12O2(g)→Cu2O(s)\mathrm{2Cu(s) + \tfrac{1}{2}O_2(g) \rightarrow Cu_2O(s)}; ΔG⊖=−78,000\Delta G^\ominus = -78{,}000 J mol−1^{-1}

H2(g)+12O2(g)→H2O(g)\mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(g)}; ΔG⊖=−1,78,000\Delta G^\ominus = -1{,}78{,}000 J mol−1^{-1}; GG is the Gibbs energy)

Answer and worked solution

Official answer: -14.7

Target reaction.
2 Cu+HX2O→CuX2O+HX2,ΔG∘=−78000−(−178000)=+100000 J mol−1.\ce{2Cu + H2O -> Cu2O + H2},\qquad\Delta G^\circ=-78000-(-178000)=+100000\ \mathrm{J\,mol^{-1}}.

Threshold. Oxidation is just prevented when ΔG=0\Delta G=0. Cu and CuX2O\ce{Cu2O} are pure solids with unit activity:
0=ΔG∘+RTln⁡pHX2pHX2O  ⟹  ln⁡pHX2pHX2O=−1000008×1250=−10.0=\Delta G^\circ+RT\ln\frac{p_{\ce{H2}}}{p_{\ce{H2O}}}\implies\ln\frac{p_{\ce{H2}}}{p_{\ce{H2O}}}=-\frac{100000}{8\times1250}=-10.

Value. pHX2O=0.01p_{\ce{H2O}}=0.01 bar, and ln⁡0.01=−2×2.3=−4.6\ln0.01=-2\times2.3=-4.6:
ln⁡pHX2=−10−4.6=−14.6.\ln p_{\ce{H2}}=-10-4.6=-14.6.

Answer: −14.6 (key: −14.70 to −14.50)

Topics:Equilibrium and Gibbs energyGibbs energy and spontaneity
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