## Help on Problem 8.99 [ENDORSED]

ntruong2H
Posts: 50
Joined: Sat Jul 09, 2016 3:00 am
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### Help on Problem 8.99

Hi, I'm having a little difficulty solving 8.99? The problem is as follows:

Hydrochloric acid oxidizes zinc metal in a reaction that produces hydrogen gas and chloride ions. A piece of zinc metal of mass 8.5 g is dropped into an apparatus containing 800.0 mL of 0.500 M HCl (aq). If the initial temperature of the hydrochloric acid solution is 25 degrees Celsius, what is the final temperature of this solution? Assume that the density and molar heat capacity of the hydrochloric acid solution are the same as those of water and that all the heat is used to raise the temperature of the solution.

I understand that I first have to find the limiting reactant and then find the enthalpy of reaction. However, I am unsure of where to go from that point.

Ariana de Souza 4C
Posts: 99
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Help on Problem 8.99

I don't get that problem either, but can you explain why you need to find the limiting reactant and how you then find the enthalpy?

Grace_Bower_2B
Posts: 13
Joined: Wed Sep 21, 2016 2:58 pm

### Re: Help on Problem 8.99  [ENDORSED]

The first thing you need to do is setup the equation so you can calculate the enthalpy of the reaction, which is mixing sold zinc and hydrochloric acid.
2HCl (aq) + Zn (s) --> H2 (g) + ZnCl2
or 2H+ (aq) + 2Cl- (aq) + Zn (s) --> H2 (g) + 2Cl- (aq) + Zn2+ (aq)
which simplfies to 2H+ (aq) + Zn (s) --> H2 (g) + Zn2+ (aq)

Now you can find the enthalpy of this reaction using the enthalpies of formation of the products.
ΔH*rxn = Σ ΔH*f(products) - Σ ΔH*f(reactants)
ΔH*rxn = [1mol(ΔH*f(H2)) + 1mol(ΔH*f(Zn2+))] - [2mol(ΔH*f(H+)) + 1mol(ΔH*f(Zn))]
ΔH*rxn = [1mol(ΔH*f(0)) + 1mol(ΔH*f(-153.89KJ/mol))] - [2mol(ΔH*f(0)) + 1mol(ΔH*f(0))]
ΔH*rxn = -153.89KJ

Then you have to modify the equation to account for the amount of zinc and HCl available. Now (like you thought) we need to find the limiting reactant
8.5gZn / 65.38g/molZn = .13 moles of Zinc
.800LHCl/(.500mol/L) = 1.6 moles of HCl
.13molZn * (2molHCl/1molZn) = .26 moles of HCl
.26 moles is less than 1.6 moles so Zinc is your limiting reactant. Therefore, the equation of the reaction that occurs in the beaker is:
.26H+ (aq) + .13Zn (s) --> .13H2 (g) + .13Zn2+ (aq)
or .13(2H+ (aq) + Zn (s) --> H2 (g) + Zn2+ (aq))

Now you can find the enthalpy of the reaction taking into account the amount of zinc and HCl. Because you multiply the equation by .13 to get the correct masses involved, you do the same to the enthalpy.
.13(-153.89KJ) = -20KJ

We can use ΔH=mcΔT=mc(Tfinal-Tinital) to solve for the final temperature.
Tfinal = ΔH/mc + Tinital

M is the mass of the solution: 800mL of the solution (which we can assume has the same density of water) means 800g of the solution
C is the specific heat capacity: We can also assume this is the same as water 4.186J/g*C.
Tinital is the intial temperature: 25*C

Then just plug and chug:
Tfinal = ΔH/mc + Tinital = 20000J/(800g*4.186J/g*C) + 25*C = 30.97*C

Ariana de Souza 4C
Posts: 99
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Help on Problem 8.99

this is all so incredibly detailed, and bless your soul, but one question--

ΔH*rxn = [1mol(ΔH*f(H2)) + 1mol(ΔH*f(Zn2+))] - [2mol(ΔH*f(H+)) + 1mol(ΔH*f(Zn))]
ΔH*rxn = [1mol(ΔH*f(0)) + 1mol(ΔH*f(-153.89KJ/mol))] - [2mol(ΔH*f(0)) + 1mol(ΔH*f(0))]
ΔH*rxn = -153.89KJ

how come you have an enthalpy of formation for Zn 2+ and not for Zn?

Grace_Bower_2B
Posts: 13
Joined: Wed Sep 21, 2016 2:58 pm

### Re: Help on Problem 8.99

The simple answer is that I literally just went to the back of the book to the table of enthalpies and the given enthalpy for Zn was zero.

The actual answer goes back to the idea of standard states. If you look on page 17 of the course reader it explains that the standard enthalpy of an element in it's most stable form is zero. This is because we calculate the standard enthalpies of formation we use the most stable forms of it's constituent elements for the reactants (ex. O2 + Cs, graphite --> CO2). For elements already in the most stable state the most stable forms of their constituent elements is just the element itself (O2 --> O2). For Zinc, zinc's most stable form is Zn (s), so the enthalpy of formation of Zn is zero. Zn2+ is not the most stable state so it's got a nonzero enthalpy of formation.