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)+H2O(g)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 In PH2  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

2Cu(s)+12O2(g)Cu2O(s);

 G=-78,000 J mol-1 

H2(g)+12O2H2O(g);

G =-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)  2Cu(s)+12O2(g)Cu2O(s);

  Δ G0 =-78,000 J mol-1

  = -78 kJ mol-1

 (ii)H2(g)+12O2H2O(g);

  Δ G0 =-178,000 J mol-1

  = -178 kJ mol-1

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

 2Cu(g)+H2O(g)Cu2O(s)+H2(g)

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

  Now, for the above reaction

  G=G0+RTln[pHpH2O] and to prevent above the reaction.

   Δ G≥ 0

So,   G0+RTln[pHpH2O]0

 After putting the values,

 105+8×1250ln[pHpH2O]0

or  105+104ln[pHpH2O]0

 or 104(lnpH2lnpH2O)105

or lnpH210+lnpH2O  

 or lnpH210+2.3log(0.01)as(  p _{H_{2}O}=1$ %) ≥ -10-4.6

So, lnpH214.6