## Help on 4.15

Tiffany Chao 2H
Posts: 117
Joined: Fri Aug 09, 2019 12:17 am

### Help on 4.15

Can someone help explain how to do 4.15 to me? I don't understand why we have to find the limiting reagent and the tabulated enthalpies of formation.

Thanks!

Hannah Lee 2F
Posts: 117
Joined: Thu Jul 11, 2019 12:15 am

### Re: Help on 4.15

In the question, a piece of zinc metal (8.5 g) is dropped into 800.0 mL of 0.500 M HCl solution, and the reaction that occurs between HCl and Zn changes the temperature of the solution. Based on the info given, the first thing you should probably do is write out the balanced chemical equation for this reaction, which turns out to be: 2HCl(aq) + Zn(s) --> H2(g) + 2Cl-(aq) + Zn2+(aq) (the last two species can just be written as ZnCl2 as well).

1. Find LR: Since the question asks for the final temperature of the solution 8.5 g of Zn was dropped in, you need to find the amount of heat energy released or absorbed in the rxn between HCl and Zn. It's implied that the external pressure is constant/not changing so qp = ΔH. Since the masses of both reactants Zn and HCl are provided, you must determine the limiting reactant to find ΔHrxn for the given amount of reactant.

0.8000 L HCl (0.500 mol/L HCl) = 0.400 HCl (1 mol Zn / 2 mol HCl) = 0.200 mol Zn required for every 0.400 mol HCl consumed --> Zn is the LR
8.5 g Zn (1 mol/65.37 g Zn) = 0.13 mol Zn

2. Find ΔH: You can find ΔHrxn through either Hess's Law, bond enthalpies, or standard enthalpies of formation, but using the standard enthalpies of formation is probably simplest for this question. You can get the standard enthalpies of formation from the appendix, and remember that ΔH°f for elements like Zn(s) and H2(g) in their most stable state = 0.

ΔH = ΣnΔH°(formation of products) - ΣnΔH°(formation of reactants) = [-153.89 kJ/mol + 2 mol (-167.16 kJ/mol) + 0] - [2 mol(-167.16 kJ/mol) + 0] = -153.89 kJ/mol rxn. So, -153.89 kJ is the energy released for one mol of reaction.

3. Adjust ΔH to find the actual energy released by 8.5 g (0.13 mol) Zn: Since Zn is the limiting reactant, we want to find the energy released when 0.13 mol of Zn is reacted (since 0.13 mol of Zn was dropped into the solution). However, based on the chemical equation, ΔHrxn gives the amount of energy released in one round of the reaction, or when 1 mol of Zn is reacted (since the stoichiometric coefficient of Zn is 1). So, use molar ratios to cancel out the units: (-153.89 kJ / mol rxn) (1 mol rxn / 1 mol Zn) (0.13 mol Zn) = -20. kJ is the energy released by 0.13 mol Zn.

4. Find final temp using qsys = -qsurr: We can consider the zinc metal that reacted the system and the solution of HCl the surroundings, so q(zinc) = -q(solution) --> -20. kJ = -mCΔT. Since the question stated the density and molar heat capacity of HCl = density and molar heat capacity of water, and 1 g = 1 mL, 800.0 mL of HCl --> 800.0 g HCl and we can use the specific heat of liquid water for C. Make sure to convert the values so that the units are consistent across the equation (so convert between kJ and J, etc), then plug in the values.

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