## 8.99 HW problem

Cerjay_Lugtu_3B
Posts: 14
Joined: Wed Sep 21, 2016 2:58 pm

### 8.99 HW problem

After finding the limiting reactant I got confused on how the solution manual got delta H. Can someone please explain to me how the they got the heat of the reaction.

Cherry_Deng_1K
Posts: 12
Joined: Wed Sep 21, 2016 2:55 pm

### Re: 8.99 HW problem

Hello,

In this case, you will have to look up the enthalpies of formation for each component of the reaction and add them up to find the enthalpy of reaction. If you look at Appendix 2A (page A10), the enthalpies of formation are listed there. Since H2(g) is the most stable form of hydrogen and Zn(s) is a solid, their enthalpies of formation are 0. You just need to look up ΔHf for ZnCl2 and HCl. Make sure that the enthalpies of formation match up with the phase and don't forget the coefficients from your balanced equation. Substitute your values into ΔHr= Σ(ΔHf products) - Σ(ΔHf reactants) to find the enthalpy of reaction. Hope this helps!

Cerjay_Lugtu_3B
Posts: 14
Joined: Wed Sep 21, 2016 2:58 pm

### Re: 8.99 HW problem

Cherry_Deng_1K wrote:Hello,

In this case, you will have to look up the enthalpies of formation for each component of the reaction and add them up to find the enthalpy of reaction. If you look at Appendix 2A (page A10), the enthalpies of formation are listed there. Since H2(g) is the most stable form of hydrogen and Zn(s) is a solid, their enthalpies of formation are 0. You just need to look up ΔHf for ZnCl2 and HCl. Make sure that the enthalpies of formation match up with the phase and don't forget the coefficients from your balanced equation. Substitute your values into ΔHr= Σ(ΔHf products) - Σ(ΔHf reactants) to find the enthalpy of reaction. Hope this helps!

I tried that but the solution manual says that delta H is -153.89 [kJ][/mol] + 2(-167.16[kJ][/mol]) - 2(-167.16[kJ][/mol]) - 0 =-153.89[kJ][/mol]

Cherry_Deng_1K
Posts: 12
Joined: Wed Sep 21, 2016 2:55 pm

### Re: 8.99 HW problem

Sorry if my explanation was unclear. All of the enthalpies of formation that you'll need to calculate the enthalpy of reaction is given in Appendix 2A. To find the ΔHf for ZnCl2 (aq), you first find the ΔHf for Zn 2+ (aq), which is the -153.89 kj/mol. You add that with 2 times the ΔHf of Cl - (aq) since there are 2 moles of chlorine per molecule of ZnCl2, which gets you 2(-167.16 kj/mol. Finally, you subtract 2 times the ΔHf for HCl (aq), which will get you -2(-167.16 kj/mol. This will get you the enthalpy of reaction, -153.89 kj/mol. Hope this was a clearer explanation!

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