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


A) -13.2

B) 13.2

C) 14.6

D) -14.6

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$