HW #53

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Shannon Han 2B
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Joined: Fri Sep 26, 2014 2:02 pm

HW #53

Postby Shannon Han 2B » Mon Jan 12, 2015 10:17 pm

For homework problem #53, the question asks to find the final temperature of the solution and the solutions manual says that the enthalpy of reaction = -153.89 + 2(-167.16) - 2(-167.16). I'm confused where they got these numbers. I looked up the enthalpy of formation of zinc and HCl and these weren't the same numbers so I'm confused. Also, where do they get 800 g? Is this because the density of water is 1 g/mL?

Kayla Denton 1A
Posts: 106
Joined: Fri Sep 26, 2014 2:02 pm

Re: HW #53

Postby Kayla Denton 1A » Tue Jan 13, 2015 1:02 pm

Hi! Those numbers are the enthalpy of formation, you were right! Make sure you look up Zn2+(aq), not Zn(s) (in appendix 2A of the textbook). As for HCl (aq), you don't need to find the number; just know that the enthalpies of formation cancel out. And you're right about the second part too; 800 g is due to the density of water = density of aqueous solution = 1 g/mL.

Nicole Leppi 2K
Posts: 20
Joined: Fri Sep 26, 2014 2:02 pm

Re: HW #53

Postby Nicole Leppi 2K » Tue Jan 13, 2015 5:15 pm

why do the enthalpies of formation cancel out? I am confused about that part.

Nicole Leppi 2K
Posts: 20
Joined: Fri Sep 26, 2014 2:02 pm

Re: HW #53

Postby Nicole Leppi 2K » Tue Jan 13, 2015 5:19 pm

Also why would you look up Zn2+ if in the solutions manual it has Zn (s) not aqueous?

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

Re: HW #53

Postby Niharika Reddy 1D » Tue Jan 13, 2015 5:39 pm

The Zn2+(aq) comes from the product ZnCl2(aq), which can be seen as Zn2+(aq) and 2Cl-(aq). The enthalpy of formation of the solid zinc is zero since zinc is in its most stable state. The enthalpy of formation of 2 moles of HCl(aq) cancels out with the enthalpy of formation of 2 moles of chloride ions since they have the same value and we subtract enthalpy of formation of reactants from enthalpy of formation of products.


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