Question 8.99

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Question 8.99

Postby JukaKim_1D » Mon Jan 18, 2016 2:17 pm

Where can I find the standard enthalpies of the reactions for HCl and ZnCl? How do I find the standard enthalpy of reaction for the equation, 2HCl (aq) + Zn(s) --> H2(g) + ZnCl2 (aq)?

Connor Olsen 3C
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Joined: Fri Sep 25, 2015 3:00 am

Re: Question 8.99

Postby Connor Olsen 3C » Mon Jan 18, 2016 5:09 pm

I found the standard enthalpy of formation for these molecules on wikipedia (

ZnCl2: -415.1
H2: 0
HCl: -167.2
Zn: 130.7

Which would thus give us the standard enthalpy of formation:
∆Hr= 0+(-415.1)-2(-167.2)-(130.7)=-211.4

However, I am confused because these numbers are very different from the ones given in the answer book... Please help if you have any further insights for this problem.

Jasmine Holloway 1G
Posts: 39
Joined: Fri Sep 25, 2015 3:00 am

Re: Question 8.99

Postby Jasmine Holloway 1G » Mon Jan 18, 2016 7:54 pm

Chem_Mod wrote: they are finding the enthalpy of reaction for the thermochemical equation using enthalpy of formations. See appendix 2A for the enthalpies of formation for the products and reactants you are looking for. HCl is listed, but for zinc chloride, you know it will dissociate into 1 mol Zn2+ (aq) and 2 mol cl- (aq), these values are listed.

To find enthalpy of rxn using enthalpy of formations, use
enthalpy of rxn=
sum enthalpy of formation products - sum enthalpy of formation reactants, being careful to multiply by moles as units are in kJ/mol.

So 1 mol Zn2+( -153.89 kJ/mol Zn2+) + 2 mol cl- (-167.16 kJ/mol cl-) + 0 - ( 2 mol HCl (-167.16 kJ/mol HCl) +0 ) = -153.89 kJ

Because you are interested in the enthalpy/mol Zinc, and 1 mol is consumed in the thermochemical rxn, then you start the next part of the problem using -153.89 kJ/mol Zn

This is what I found on another post. Hope this helped.

Sophia 3F
Posts: 17
Joined: Fri Sep 25, 2015 3:00 am

Re: Question 8.99

Postby Sophia 3F » Fri Jan 22, 2016 1:59 pm

I'm a bit confused on how it was solved in the solutions manual. I thought enthalpy of reaction= (sum of formations of products)-(sum of formations of reactants). In the solutions manual though they did (sum of formations of reactants)-(sum of formations of products)?

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