Discussion Forum

1)

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 vapor as an impurity. The water vapor oxidizes copper as per the reaction is given below

$2Cu (g)  + H_{2}O (g) \rightarrow Cu_{2}O (s)+H_{2}(g)$

  $P_{H_{2}}$   is the minimum partial  pressure of H₂ (in bar) needed to prevent the oxidation at 1250 K. The value of In $P_{H_{2}}$  is........

[Given : total pressure = 1 bar , R  ( universal gas constant ) = 8 JK^-1 mol^-1 , ln(10)=2.30 Cu(s)  and Cu₂ O(s) are  mutually immiscible.)

  At 1250 K

$2 Cu (s)+\frac{1}{2}O_{2}(g)\rightarrow Cu_{2}O (s);$

 $\triangle G^{\ominus} $=-78,000 J mol^-1 

$H_{2}(g)+\frac{1}{2}O_{2}\rightarrow H_{2}O (g);$

$\triangle G^{\ominus}$ =-178,000 J mol^-1

  G  is the Gibs energy

Answer:

Option D

Explanation:

Given

(i)  $2 Cu (s)+\frac{1}{2}O_{2}(g)\rightarrow Cu_{2}O (s);$

  Δ G⁰ =-78,000 J mol^-1

  = -78 kJ mol^-1

 (ii)$H_{2}(g)+\frac{1}{2}O_{2}\rightarrow H_{2}O (g);$

  Δ G⁰ =-178,000 J mol^-1

  = -178 kJ mol^-1

   So, net reaction  is (By (i) - (ii))

 $2Cu (g)  + H_{2}O (g) \rightarrow Cu_{2}O (s)+H_{2}(g)$

  Δ G = 100000 J /mol or 10⁵ J/mol=100 kJ mol^-1

  Now, for the above reaction

  $\triangle G = \triangle G^{0}+RT ln[\frac{p_{H_{}}}{p_{H_{2}O}}]$ and to prevent above the reaction.

   Δ G≥ 0

So,   $\triangle G^{0}+RT ln[\frac{p_{H_{}}}{p_{H_{2}O}}] \geq 0$

 After putting the values,

 $10^{5}+8\times1250  ln[\frac{p_{H_{}}}{p_{H_{2}O}}] \geq 0$

or  $10^{5}+10^{4}  ln[\frac{p_{H_{}}}{p_{H_{2}O}}] \geq 0$

 or $10^{4}  (lnp _{H_{2}}-ln p_{H_{2}O})\geq -10^{5}$

or $lnp _{H_{2}} \geq -10 +ln p_{H_{2}O}$  

 or $lnp _{H_{2}} \geq -10 +2.3log(0.01) as  ( $  p _{H_{2}O}=1$ %) ≥ -10-4.6

So, $ lnp _{H_{2}}\geq-14.6$