JEE Challenger
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ChemHardAdv 2018 • Paper 2

Minimum Partial Pressure of Hydrogen to Prevent Copper Oxidation

The copper surface undergoes tarnishing via oxidation to copper oxide. In an experiment, N2\text{N}_2 gas stream is passed to prevent the oxide formation during the heating of copper at 1250 K1250\text{ K}. However, this N2\text{N}_2 gas contains 1 mole %1\text{ mole }\% of water vapour as an impurity. The water vapour oxidises copper via the following reaction: 2Cu(s)+H2O(g)→Cu2O(s)+H2(g)2\text{Cu(s)} + \text{H}_2\text{O(g)} \rightarrow \text{Cu}_2\text{O(s)} + \text{H}_2\text{(g)} Let pH2p_{\text{H}_2} denote the minimum partial pressure of H2\text{H}_2 (in bar) needed to inhibit the oxidation at 1250 K1250\text{ K}. Determine the value of ln⁡(pH2)\ln(p_{\text{H}_2}).

(Given: total pressure =1 bar= 1\text{ bar}, R (universal gas constant)=8 J K−1mol−1R \text{ (universal gas constant)} = 8\text{ J K}^{-1}\text{mol}^{-1}, ln⁡(10)=2.3\ln(10) = 2.3. Cu(s)\text{Cu(s)} and Cu2O(s)\text{Cu}_2\text{O(s)} are mutually immiscible. At 1250 K1250\text{ K}: 2Cu(s)+12O2(g)→Cu2O(s);ΔG⊖=−78,000 J mol−12\text{Cu(s)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{Cu}_2\text{O(s)}; \quad \Delta G^\ominus = -78,000\text{ J mol}^{-1} H2(g)+12O2(g)→H2O(g);ΔG⊖=−1,78,000 J mol−1;\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{H}_2\text{O(g)}; \quad \Delta G^\ominus = -1,78,000\text{ J mol}^{-1}; where GG denotes the Gibbs energy)

Official Numerical Answer-14.6

Topics & Concepts

ThermodynamicsGibbs Free EnergyChemical Equilibrium

Step-by-Step Solution

To find the minimum partial pressure of H2\text{H}_2 (pH2p_{\text{H}_2}) required to inhibit the oxidation of copper at 1250 K1250\text{ K}, we consider the given chemical reactions:

  1. Oxidation of Copper: 2Cu(s)+12O2(g)→Cu2O(s),ΔG1⊖=−78,000 J mol−12\text{Cu(s)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{Cu}_2\text{O(s)}, \quad \Delta G_1^\ominus = -78,000\text{ J mol}^{-1}

  2. Oxidation of Hydrogen: H2(g)+12O2(g)→H2O(g),ΔG2⊖=−1,78,000 J mol−1\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{H}_2\text{O(g)}, \quad \Delta G_2^\ominus = -1,78,000\text{ J mol}^{-1}

Subtracting reaction (2) from reaction (1) gives the target reaction: 2Cu(s)+H2O(g)⇌Cu2O(s)+H2(g)2\text{Cu(s)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{Cu}_2\text{O(s)} + \text{H}_2\text{(g)}

The standard Gibbs free energy change ΔG⊖\Delta G^\ominus for this reaction is: ΔG⊖=ΔG1⊖−ΔG2⊖=−78,000−(−1,78,000)=+1,00,000 J mol−1\Delta G^\ominus = \Delta G_1^\ominus - \Delta G_2^\ominus = -78,000 - (-1,78,000) = +1,00,000\text{ J mol}^{-1}

The equilibrium constant KpK_p for the reaction is given by: ΔG⊖=−RTln⁡Kp\Delta G^\ominus = -RT \ln K_p

Given:

  • R=8 J K−1mol−1R = 8\text{ J K}^{-1}\text{mol}^{-1}
  • T=1250 KT = 1250\text{ K}

Thus, RT=8×1250=10,000 J mol−1RT = 8 \times 1250 = 10,000\text{ J mol}^{-1}

ln⁡Kp=−ΔG⊖RT=−1,00,00010,000=−10\ln K_p = -\frac{\Delta G^\ominus}{RT} = -\frac{1,00,000}{10,000} = -10

Since Cu(s)\text{Cu(s)} and Cu2O(s)\text{Cu}_2\text{O(s)} are pure mutually immiscible solids, their activities are unity (a=1a = 1). Therefore, the equilibrium expression in terms of partial pressures is: Kp=pH2pH2OK_p = \frac{p_{\text{H}_2}}{p_{\text{H}_2\text{O}}}

To inhibit the oxidation reaction (prevent the forward process), the reaction quotient QQ must satisfy: Q≥Kp  ⟹  pH2pH2O≥KpQ \ge K_p \implies \frac{p_{\text{H}_2}}{p_{\text{H}_2\text{O}}} \ge K_p

Hence, the minimum partial pressure of H2\text{H}_2 corresponds to the equilibrium state: pH2=Kp⋅pH2Op_{\text{H}_2} = K_p \cdot p_{\text{H}_2\text{O}}

Given that the total pressure is 1 bar1\text{ bar} and the gas contains 1 mole %1\text{ mole }\% H2O\text{H}_2\text{O}: pH2O=(1100)×1 bar=10−2 barp_{\text{H}_2\text{O}} = \left(\frac{1}{100}\right) \times 1\text{ bar} = 10^{-2}\text{ bar}

Taking the natural logarithm of both sides: ln⁡(pH2)=ln⁡Kp+ln⁡(pH2O)\ln(p_{\text{H}_2}) = \ln K_p + \ln(p_{\text{H}_2\text{O}}) ln⁡(pH2)=−10+ln⁡(10−2)=−10−2ln⁡(10)\ln(p_{\text{H}_2}) = -10 + \ln(10^{-2}) = -10 - 2\ln(10)

Using the given value ln⁡(10)=2.3\ln(10) = 2.3: ln⁡(pH2)=−10−2(2.3)=−10−4.6=−14.6\ln(p_{\text{H}_2}) = -10 - 2(2.3) = -10 - 4.6 = -14.6

Final Answer: The value of ln⁡(pH2)\ln(p_{\text{H}_2}) is −14.6-14.6.

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