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JEE Advanced · Chemistry · 6. Thermodynamics (C)

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

2Cus+ H2Og Cu2O(s) + H2(g)

PH2 Is the minimum partial pressure of H2 (in bar) needed to prevent the oxidation at 1250 K. The value of lnpH2 is ____.

(Given: total pressure \(=1\) bar, R (universal gas constant) \(=8 J K ^{-1} mol^{-1}, \ln (10) =2.3 \cdot Cu (s)\) and \(Cu _2 O (s)\) are mutually immiscible.

At \(1250 K: 2 Cu ( s )+\frac{1}{2} O _2(g) \rightarrow Cu _2 O ( s ) ;\) \(\Delta G^{\ominus}=-78,000 J mol ^{-1}\)

\(H _2(g)+\frac{1}{2} O _2(g) \rightarrow H _2 O ( g ) ; \quad \Delta^{\ominus}=\)\(-1,78,000 J mol ^{-1} ; G\) is the Gibbs energy)/mi> \(\Theta=\) \(-78,000 J mol -1\)

\(H _2(g)+\frac{1}{2} O _2(g) \rightarrow H _2 O ( g ) ; \quad\)\(\Delta^{\ominus}=-1,78,000 J mol ^{-1} ; G\) is the Gibbs energy)

  1. A 14.6
  2. B 14.52
  3. C 14.55
  4. D 14.2
Verified Solution

Answer & Solution

Correct Answer

(A) 14.6

Step-by-step Solution

Detailed explanation

\(2 Cu (s)+\frac{1}{4} O _2(g) \rightarrow 1 Cu _2 O (s)\) \(\Delta G^0=-78 k J\)
\(\left[ H _2(g)+\frac{1}{2} O _2 \rightarrow H _2 O (g)\right.\) \(\left.\Delta G^o=-178 \quad k J\right] \times(-1)\)
Hence, \(2 Cu ( s )+ H _2 O ( g ) \rightarrow Cu _2 O + H _2(g)\) \(\Delta G^o=+100 k J\)
\(\Delta G=\Delta G^o+R T \ln Q\)
\(0=+100+\frac{8}{1000} \times 1250 \ln \frac{P_{H_2}}{P H_2 O}\)
\(-\frac{100 \times 1000}{8}=1250 \ln \frac{P_{H 2}}{\left(\frac{1}{100} \times 1\right)}\)
\(\ln P_{H_2}=-14.6\)
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