1.since for both the reaction, deltaG is -ve, the reactions are spontaneous.
2. for the reaction Pb(s)+CO(g) ------>PbO(s)+C(graphite)
for solids in the elemental state, deltaG=0 hence for graphite and Pb, detlaG=0
deltaG for the reaction = deltaG of products- detlaG of reactants = 1* deltaG of PbO+1* deltaG of C- {1* deltaG of Pb+ 1* deltaG of CO) = -51
deltaG of CO=-274/2=-137
hence deltaG of PbO=-51-137= -188 Kj/mole
hence the third statemt is incorrect.
for the reaction-2 from deltaG= deltaH-T*deltaS
deltaS= (deltaH- deltaG)/T = (-221+274)*1000/298.15=178 J/mole.K so third is also incorrect.
Reaction is spontaneous at all temperatures since deltaH and deltaS are independent of temperature
deltaS for the reaction -2 = 2*entropy of CO -(2* entroyp of C +1*entroy of O2}=2*197.9-(2*5.69+205)=179.42 J/mole.K
deltaG at 500K= deltaH-T*deltaS= -221-179.42*500/1000 Kj=-Ve, the reaction is spontaneous at 500K as well.
So statements 1 and 3 are incorrect.
15. Consider the following two reactions, with thermodynamic data at 298.15 K (1) Pb(s) + CO(g)...
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