1.since for both the reaction, deltaG is -ve, the reactions are spontaneous. 2.

Question

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.

Explanation / Answer

15. Consider the following two reactions, with thermodynamic data at 298.15 K (1) Pb(s) + CO(g) Pb0(s) + C(graphite) (2) 2 C(graphite) + O2(g) 2 CO(g) Ar =-107 kl mori G. =-51 kl mor Which of the following statements is(are) FALSE? (Assume AHt and AS" are independent of temperature.) G,-[PbO(s)) +188 kJ mol-1 Both reactions are spontaneous under standard conditions at room temperature. AS"for reaction 2 is-178 J K-1 mori at 298.15 K. Reaction 2 is spontaneous at 500°C when the partial pressures of both O2 and CO are 1 bar. (iii) (iv) B) iii C) i E) iv

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