Question

Can someone help me solve this and explain it?

3. (12 pts) You want to calculate AH for the following reaction: 4 PbCO3(s) 4 Pb(s)+ 4 C(s)+6 O2(g) ΔΗ-? You know 1 Pb(s) +½ O2(g) PbO(s) 2 C(s) +2 O2(g) 2 CO2(g) PBCO3(s) PbO(s) + CO2(g) AH438 kJ AH=-788 kJ AH=86 kJ What is AH for the desired reaction? You must show all work to get full credit. (You can make sure you get the correct answer by showing that you get the desired reaction)

Answer 1

$\Delta H=+3672kJ$ for the desired reaction.

The reactions are-

1 Pb(s) O2(g) PbO(s) 2 C(s) 2 O2(g) 2 CO2(g) PbCO3(s) PbO(s) + CO2(g) AH438 kJ 1 AH788 kJ 2 AH 86 kJ

4 Рьсоз с ) 4РЬСS) t 4C9t 6 020я) + 4с 4 Рьсоз (9) D4РЬО Х4 Ан -+86 хч %3D f844 RI 4909 с9) 4со. 4 С с5) + 402g) - ++8 x2 = +1516 RJ. L Дн X 2 НРЬО + 202гg) .4РЬСЯ) ху 152 k дн - +438х4 + 4Рbrs) сOзерт - 4 Gs, 4 PЬсоз о) АН 3+3612 ET
Here directions of reactions 1 and 2 are reversed, so their signs of $\Delta H$ are also reversed and multiplied with proper coefficients to balance the overall reaction.