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Thermodynamics question

2018 · Shift 2 · Q8
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Thermodynamics question

2018 · Shift 2 · Q8

JEE AdvancedChemistryThermodynamicsNumerical+3 / −1
The surface of copper gets tarnished by the formation of copper oxide. N2{N_2}N2​ gas was passed to prevent the oxide formation during heating of copper at 1250K.1250K.1250K. However, the N2{N_2}N2​ gas contains 111 mole % of water vapor as impurity. The water vapor oxidises copper as per the reaction given below : 2Cu(s)+H2O(g)→Cu2O(s)+H2(g)2Cu\left( s \right) + {H_2}O\left( g \right) \to C{u_2}O\left( s \right) + {H_2}\left( g \right)2Cu(s)+H2​O(g)→Cu2​O(s)+H2​(g)PH2{P_{H2}}PH2​ is the minimum partial pressure of H2{H_2}H2​(in bar) needed to prevent the oxidation at 1250K.1250K.1250K. The value of ln⁡(PH2)\ln \left( {{P_{H2}}} \right)ln(PH2​) is ‾\underline{\hspace{2cm}}​. Given: total pressure =1=1=1 bar, RRR(universal gas constant ) =8JK−1  mol−1,=8J{K^{ - 1}}\,\,mo{l^{ - 1}},=8JK−1mol−1,ln⁡(10)=2.3. \ln \left( {10} \right) = 2.3.\,ln(10)=2.3.Cu(s)Cu(s)Cu(s) and Cu2O(s)C{u_2}O\left( s \right)Cu2​O(s) are naturally immiscible. At 1250K:2Cu(s)1250K:2Cu(s)1250K:2Cu(s)+12O2(g)→Cu2O(s);+ \frac{1}{2}{O_2}\left( g \right) \to C{u_2}O\left( s \right);+21​O2​(g)→Cu2​O(s);ΔG∘=−78,000J mol−1\Delta {G^ \circ } = - 78,000J\,mo{l^{ - 1}}ΔG∘=−78,000Jmol−1H2(g)+12O2(g)→H2O(g);{H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( g \right);H2​(g)+21​O2​(g)→H2​O(g);                  ΔG∘=−1,78,000J mol−1;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {G^ \circ } = - 1,78,000J\,mo{l^{ - 1}};ΔG∘=−1,78,000Jmol−1; (GGG is the Gibbs energy)
Numerical answer
View written solutionFree

Correct answer: -14.6

  1. Required condition to prevent oxidation

The reaction causing oxidation is 2Cu(s)+H2O(g)→Cu2O(s)+H2(g)2Cu(s)+H_2O(g)\rightarrow Cu_2O(s)+H_2(g)2Cu(s)+H2​O(g)→Cu2​O(s)+H2​(g)

To just prevent oxidation, this reaction must be at equilibrium: ΔG=0\Delta G=0ΔG=0

Since solids have activity =1=1=1, Q=PH2PH2OQ=\frac{P_{H_2}}{P_{H_2O}}Q=PH2​O​PH2​​​

So at equilibrium, K=PH2PH2OK=\frac{P_{H_2}}{P_{H_2O}}K=PH2​O​PH2​​​

Thus the minimum hydrogen partial pressure is PH2=K PH2OP_{H_2}=K\,P_{H_2O}PH2​​=KPH2​O​

  1. Find standard Gibbs energy for the given reaction

Given: 2Cu(s)+12O2(g)→Cu2O(s),ΔG1∘=−78000 J mol−12Cu(s)+\frac12 O_2(g)\rightarrow Cu_2O(s),\qquad \Delta G_1^\circ=-78000\text{ J mol}^{-1}2Cu(s)+21​O2​(g)→Cu2​O(s),ΔG1∘​=−78000 J mol−1 H2(g)+12O2(g)→H2O(g),ΔG2∘=−178000 J mol−1H_2(g)+\frac12 O_2(g)\rightarrow H_2O(g),\qquad \Delta G_2^\circ=-178000\text{ J mol}^{-1}H2​(g)+21​O2​(g)→H2​O(g),ΔG2∘​=−178000 J mol−1

We need: 2Cu(s)+H2O(g)→Cu2O(s)+H2(g)2Cu(s)+H_2O(g)\rightarrow Cu_2O(s)+H_2(g)2Cu(s)+H2​O(g)→Cu2​O(s)+H2​(g)

This can be obtained as:

  • Reaction (1) as it is
  • plus reverse of reaction (2): H2O(g)→H2(g)+12O2(g),ΔG∘=+178000 J mol−1H_2O(g)\rightarrow H_2(g)+\frac12 O_2(g),\qquad \Delta G^\circ=+178000\text{ J mol}^{-1}H2​O(g)→H2​(g)+21​O2​(g),ΔG∘=+178000 J mol−1

Adding, 2Cu(s)+H2O(g)→Cu2O(s)+H2(g)2Cu(s)+H_2O(g)\rightarrow Cu_2O(s)+H_2(g)2Cu(s)+H2​O(g)→Cu2​O(s)+H2​(g)

Hence, ΔG∘=−78000+178000=100000 J mol−1\Delta G^\circ=-78000+178000=100000\text{ J mol}^{-1}ΔG∘=−78000+178000=100000 J mol−1

  1. Calculate equilibrium constant

Using ΔG∘=−RTln⁡K\Delta G^\circ=-RT\ln KΔG∘=−RTlnK

So,

=-\frac{100000}{8\times1250} =-\frac{100000}{10000}=-10$$ Thus, $$K=e^{-10}$$ 4. **Use water impurity information** Nitrogen contains $1$ mole % water vapour at total pressure $1$ bar. Therefore, $$P_{H_2O}=0.01\text{ bar}=10^{-2}\text{ bar}$$ At the threshold of oxidation prevention, $$P_{H_2}=K P_{H_2O}=e^{-10}\times10^{-2}$$ Taking natural log, $$\ln P_{H_2}=\ln K+\ln(10^{-2})$$ $$\ln P_{H_2}=-10-2\ln 10$$ Given $\ln 10=2.3$, $$\ln P_{H_2}=-10-2(2.3)=-10-4.6=-14.6$$ 5. **Final answer** $$\boxed{\ln(P_{H_2})=-14.6}$$
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