Question

calculate the standard free energy change for the following reaction Pb(l) + 1/2 O2(g) = PbO(s) at 527 C

(EBB 160 -7- [b] Calculate the standard free energy change for the following reaction Pb (I)+ 2 O2 (g)-PbO (s) at 527°C given the following data AH 298 PhOUS)-220,080 J/mol AH R00 PROUS 220,424,4-JHmot 8 k ASo 298,PbO(S)=68.04 JIKImol AS 298,Ph(8)65.1 JIKImol 205.88 J/KImol AS 298,02(g) SG.Cn /K/mol CP PHOS)=44.52 +16.8x10 JIK/mol V Cp.Phe)=23.65+9.79x10( K/mol Cp Pory=32.55-3.108x10 1.68x10% Croe)=30.07+4.2x10 JIKImol 327°C Melting point of Pb, Tm Latent heat of fusion, Pb, L 4830 J/mol (60 marks/ma Define a catalyst [a] Takrifkan pemangkin ubcione thai con dier he rote or bu eman nchanqed (15 marks/m o omouni a the end HaO ev2H.0. 06

Answer 1

Phols) Pb(L) 02 C9) O-TOs we Knous thet and oo becaur they ar SHPh0 in ptandard stte Bute her O 220080J/m) os0P - Pb) s(0) 2 205 88 x 2 65.1 ニ 68,0y -100T/K T=C537+R73) g00K Value put All the A2-2 2 0 080 -3 1丁210 Kナ = -140080J/mol