JEE Advanced · Chemistry · 6. Thermodynamics (C)
The surface of copper gets tarnished by the formation of copper oxide. gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:
Is the minimum partial pressure of (in bar) needed to prevent the oxidation at 1250 K. The value of 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)
- A 14.6
- B 14.52
- C 14.55
- D 14.2
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, \(2 Cu ( s )+ H _2 O ( g ) \rightarrow Cu _2 O + H _2(g)\) \(\Delta G^o=+100 k J\)
\(\Delta G=\Delta G^o+R T \ln Q\)
\(0=+100+\frac{8}{1000} \times 1250 \ln \frac{P_{H_2}}{P H_2 O}\)
\(-\frac{100 \times 1000}{8}=1250 \ln \frac{P_{H 2}}{\left(\frac{1}{100} \times 1\right)}\)
\(\ln P_{H_2}=-14.6\)
\(\left[ H _2(g)+\frac{1}{2} O _2 \rightarrow H _2 O (g)\right.\) \(\left.\Delta G^o=-178 \quad k J\right] \times(-1)\)
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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