Focus 4 Exercise 15

[Talia Dini - 3I]

Hi! Could someone please explain how they solved this problem?

4.15 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 8C, 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.

[Sophia Hu 1A]

First I wrote out the balanced equation:
2 HCl (aq) + Zn (s) --> H2 (g) + ZnCl2(aq)
ZnCl2 (aq) is the same thing as Zn2+ and 2Cl- because they dissociate into their ions

Then we have to find the limiting reagent:
HCl: (0.500 mol/L)(0.8000 L) = 0.4 mol HCl
Zn: (8.5 g) / (65.39 g/mol) = 0.13 mol Zn

0.4 mol HCl (1 mol Zn/2 mol HCl) = 0.2 mol Zn required
Since there is only 0.13 mol Zn added then Zn is the limiting reagent. This will help us after we find the enthalpy change of the reaction.

Remember deltaHrxn = sum(deltaHf products) - sum(deltaHf reactants)
You can find the heat of formation values in Appendix 2 at the end of the textbook
We know that ZnCl2 dissociates into its ion
Zn2+: deltaHf = - 153.89 kJ/mol
HCl: deltaHf = - 167.16 kJ/mol
Cl-: deltaHf = - 167.16 kJ/mol
deltaHrxn = [1 mol(- 153.89 kJ/mol) + 2mol(- 167.16 kJ/mol)] - [2 mol(- 167.16 kJ/mol)] = - 153.89 kJ/mol

The deltaHrxn can be thought of as - 153.89 kJ per mol Zn, so we can use the actual amount of Zn added to find the enthalpy change of the reaction: - 153.89 kJ/mol (0.13 mol) = - 20.0 kJ

The density of water is 1 g/mL, so 800.0 mL (1 g/mol) = 800.0 g
q = mcdeltaT
deltaT = q/mc = (- 20.0 x 10^3 J)/[(800.0 g)(4.184 J)] = 5.98 C
deltaT = Tf - Ti and we know Ti = 25 C
Tf = deltaT + Ti = 5.98 + 25 = 31 C

[SophiaNguyen_2L]

I think I'm most confused by the deltaHrxn here, especially those segment of the calculations: 2mol(- 167.16 kJ/mol). Where do you get the 2 from? And why do you multiply it by what I'm assuming is the deltaHf of Cl-? Thank you!

[Brittney Nguyen 2L]

I think I'm most confused by the deltaHrxn here, especially those segment of the calculations: 2mol(- 167.16 kJ/mol). Where do you get the 2 from? And why do you multiply it by what I'm assuming is the deltaHf of Cl-? Thank you!

Hi!

The 2 comes from the balanced reaction. I can rewrite the reaction more explicitly for you:
2HCl (aq) + Zn (s) -> H2 (g) + Zn^(2+) (aq) + 2Cl^- (aq)

We know that Zn (s) and H2 (g) are in their standard states, so their standard enthalpies of formation are both 0 kJ/mol.

The standard enthalpies of formation of HCl and Cl- are both -167.16 kJ/mol, and the standard enthalpy of formation of Zn^(2+) is -153.89 kJ/mol.

So the 2mol(-167.16 kJ/mol) can actually refer to both the HCl and the Cl-, as two of each are present in the balanced equation, as Sophia noted.

I hope this helps!

[Steph Du 1H]

First I wrote out the balanced equation:
2 HCl (aq) + Zn (s) --> H2 (g) + ZnCl2(aq)
ZnCl2 (aq) is the same thing as Zn2+ and 2Cl- because they dissociate into their ions

Then we have to find the limiting reagent:
HCl: (0.500 mol/L)(0.8000 L) = 0.4 mol HCl
Zn: (8.5 g) / (65.39 g/mol) = 0.13 mol Zn

0.4 mol HCl (1 mol Zn/2 mol HCl) = 0.2 mol Zn required
Since there is only 0.13 mol Zn added then Zn is the limiting reagent. This will help us after we find the enthalpy change of the reaction.

Remember deltaHrxn = sum(deltaHf products) - sum(deltaHf reactants)
You can find the heat of formation values in Appendix 2 at the end of the textbook
We know that ZnCl2 dissociates into its ion
Zn2+: deltaHf = - 153.89 kJ/mol
HCl: deltaHf = - 167.16 kJ/mol
Cl-: deltaHf = - 167.16 kJ/mol
deltaHrxn = [1 mol(- 153.89 kJ/mol) + 2mol(- 167.16 kJ/mol)] - [2 mol(- 167.16 kJ/mol)] = - 153.89 kJ/mol

The deltaHrxn can be thought of as - 153.89 kJ per mol Zn, so we can use the actual amount of Zn added to find the enthalpy change of the reaction: - 153.89 kJ/mol (0.13 mol) = - 20.0 kJ

The density of water is 1 g/mL, so 800.0 mL (1 g/mol) = 800.0 g
q = mcdeltaT
deltaT = q/mc = (- 20.0 x 10^3 J)/[(800.0 g)(4.184 J)] = 5.98 C
deltaT = Tf - Ti and we know Ti = 25 C
Tf = deltaT + Ti = 5.98 + 25 = 31 C

Why is the deltaT positive when the q is negative? In the answer key, they put -4.184 so that it becomes positive. How would you know to do this?

[Sophia Hu 1A]

Hi I am so sorry for the late response!

The reason the deltaT is positive is because the zinc is dropped into the hydrochloric acid solution. The reaction between the zinc and the hydrochloric acid produces heat. Because the heat released is used to heat up the solution then the change in temperature should be positive.

Another way to set up this problem is writing - q_rxn = q_solution

We can use q = n*deltaH and q = mc*deltaT
For the reaction, we found the amount of heat released is deltaH = - 153.89 kJ per mol of Zinc

Therefore, - (153.89 x 10^3 kJ/1 molZn)(8.5 g Zn * 1mol/65.39g) = 800.0 g (4.184 J/C*g)(Tf - 25)
When you solve for Tf, you get 31C

I hope this helps! Sorry again this is so late