Question #3b059

1 Answer
Al E.
Dec 28, 2017

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.

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