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)+12O2→H2O(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(lnpH2−lnpH2O)≥−105
or lnpH2≥−10+lnpH2O
or lnpH2≥−10+2.3log(0.01)as( p _{H_{2}O}=1$ %) ≥ -10-4.6
So, lnpH2≥−14.6