Question

Pb(s) + PbO2(s)+2SO3(g)——>2PbSO4(s)
Calculate the delta H for the reaction below at 25 degrees celcius

Answer 1

$\Delta H^0_r=$

$=2*\Delta H^0_{f.PbSO_4(s.)} - 2*\Delta H^0_{f.SO_3(g.)}+$

$+ (-\Delta H^0_{f.PbO_2(s)}) -\Delta H^0_{f.Pb(s.)}$

$\Delta H^0_{f.Pb(s.)} = 0 \ kJ/mol$

$\Delta H^0_{f.PbO_2(s)} = -277.4\ kJ/mol$

$\Delta H^0_{f.SO_3(g.)} = -395.72\ kJ/mol$

$\Delta H^0_{f.PbSO_4(s.)} = -919.94 \ kJ/mol$

$\Delta H^0_r= (2*-919.94) - (2*-395.72) + 277.4 =$

$= -771.04\ kJ/mol$

Answer: $\Delta H^0_r=-771.04\ kJ/mol$

All values were taken from https://chem.pg.edu.pl/documents/175260/14212622/chf_epm_cr_00.pdf