Enthalpy of formation of Zinc

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Chem_Mod
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Enthalpy of formation of Zinc

Postby Chem_Mod » Tue May 14, 2013 4:33 pm

We have to use the enthalpy of formation to find the energy required to oxidize zinc, however I'm unable to find the value for the formation of ZnCl2. How are we supposed to find this?

Chem_Mod
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Re: Enthalpy of formation of Zinc

Postby Chem_Mod » Tue May 14, 2013 4:33 pm

Appendix 2 on page A15 in the back of the book has the enthalpy of formation for the Zn+2 ion which = -153.89 kJ/mol and is the value needed to solve the problem.

Quynh Nguyen 2O
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Re: Enthalpy of formation of Zinc

Postby Quynh Nguyen 2O » Mon Jan 19, 2015 3:24 pm

Why don't we account for Cl2 in ZnCl2?

Also, this question refers to the problem 7.53 in the textbook and it asks for the Delta Hrxn using standard enthalpy formation.

2HCl + Zn ---> H2 + ZnCl2

Delta Hrxn = Hf (ZnCl2) - Hf 2(HCl)
Since we don't include Zn and H2.

But why is that wrong according to the solution manual?

Niharika Reddy 1D
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Re: Enthalpy of formation of Zinc

Postby Niharika Reddy 1D » Mon Jan 19, 2015 3:41 pm

2HCl(aq)+ Zn(s) ---> H2(g) + ZnCl2(aq)
Since HCl is a strong acid, it dissociates and we can write it as 2H+(aq)+2Cl-(aq). ZnCl2 is aqueous and soluble so we can also write that in its dissociated form: Zn2+(aq)+2Cl-(aq). Substitute the dissociated forms into the initial reaction to give the complete ionic equation, and cancel species that appear on both sides (spectator ions) to get the net ionic equation. Since 2 moles of chloride ions appear on both sides, they cancel out. This results in:
2H+(aq)+Zn(s)---> H2(g)+Zn2+(aq)
ΔHrxn is the heat of formation of Zn2+(aq) only, since all the other species have zero as their heat of formation.


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