Question

Question #3b059

Answer 1

With the reaction,

`Pb(s) + PbO_2(s) to 2PbO(g)`

we're asked to determine its enthalpy change. Generally this is accomplished by subtracting the sum of enthalpy of formation of reactants from products.

Hence,

`DeltaH^° =(2mol * (-219.0kJ)/(mol)) - (0 + 1mol * (-276.6kJ)/(mol)) approx -161.4kJ`

This reaction is considered exothermic. It releases heat as it progresses.