Question

Zinc and copper (I) oxide react exothermically to produce copper and zinc oxide. Use Hess’s law to calculate the enthalpy of this reaction from the heats of formation of copper (I) oxide and zinc oxide?. Find the necessary information on the Internet to solve this problem.

Answer 1

Balanced equation is

Zn(s) + Cu2O ------> 2Cu(s) + ZnO(s)

According to Hess's law of heat of summation

∆H°rxn = ∆H°f(products) - ∆H°f(reactants)

∆H°f(products) = (2×∆H°f(Cu(s)) + (1×∆H°f (ZnO(s))

For elements in their standard state ∆H°f is Zero because they are not formed and just they are present

So, ∆H°f(Cu(s)) = 0

∆H°f ( ZnO(s))=-348.28kJ/mol

Therefore ,

∆H°f(products) =( 2×0 ) + ( -348.28kJ/mol)

= -348.28kJ/mol

∆H°f ( reactants)=(1×∆H°f(Zn(s)) ) + ( 1× ∆H°f(Cu2O(s)))

∆H°fZn(s) = 0

∆H°f(Cu2O(s))=-168.62kJ/mol

Therefore,

∆H°f ( reactants)= (1×0) + ( 1× (-168.62kJ/mol))

= - 168.62kJ/mol

Now, The ∆H°rxn is

∆H°rxn = (-348.28kJ/mol) - (-168.62kJ/mol)

= -179.7kJ/mol

So, the enthalpy of the reaction is -179.7kJ/mol