Question

calculate the molar enthalpy

Zn+2HCl→ZnCl2+H2 -80.7kj/mol

Zn+1/2O2→ZnO ?

ZnO​(s)​ + H+​ ​ ​→​ H​2O​ ​(l)​ + Zn2​ +​(aq) and using 1 g of solid zinc oxide (with enough acid) only gives 0.235 kJ of heat. And finally, the heat of formation of liquid water is -285.9 kJ/mol.

Answer 1

Find out the enthalpies of formation for all compounds involved in the reaction using the source given in Resources. Those values are usually given in kilojoules (kJ):
Zn = 0 kJ
HCl = -167.2 kJ
ZnCl2 = -415.1 kJ
H2 = 0 kJ
Enthalpies of formation of elements such as Zn or H2 are equal to zero.

• Add up the enthalpies of formation of the reagents of the reaction. The reagents are zinc and hydrochloric acid, and the sum is 0 + 2 * ( -167.2) = -334.3. Note that the heat of formation of HCl is multiplied by 2 because the reaction coefficient of this compound is 2.

• Sum up the enthalpies of formation of the reaction products. For this reaction, products are zinc chloride and hydrogen, and the sum is -415.1 + 0 = -415.1 kJ.

• Subtract the enthalpy of the reagents from the enthalpy of the products to calculate the enthalpy (heat) of the zinc reaction; the enthalpy is -415.1 - (-334.3) = -80.7 kJ.

The enthalpy of the reaction is (the sum of the enthalpy of products - the sum of the enthalpy of the reactants).

In this problem, are the products.

The enthalpy formation of any element (in this case is ) is 0 because gives zero.

Therefore, [2(-285.9)-0] - [(0.235) + (0) + (-80.7)] = -552.83 kJ.