## 4.15 7th ed

inlovewithchemistry
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### 4.15 7th ed

Why can you not use mc(delta)T=-mc(delta)T to solve this?? I keep getting 298 (if i use kelvin) and 25 (if i use celsius) as my answer.

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### Re: 4.15 7th ed

This question is not as simple as adding a hot piece of metal to a liquid and using qliquid= -qmetal to find the change in temperature. The information given does not tell you the initial temperature of the Zn, and you can't assume it is the same as the HCl and use qliquid= -qmetal. To solve this problem you need to first realize that the reaction you are performing will have some kind of enthalpy change, which will alter the final temperature of the HCl. So the reaction itself is causing the heat change, not the addition of a hot piece of metal. To calculate the enthalpy you first need to determine the balanced equation, which is:
$2HCl (aq) + Zn (s) \rightarrow H_{2} (g) +ZnCl_{2}$

Using this and the quantities of each reactant given you need to determine the limiting reagent. With 0.8L of 0.5 M HCl there are 0.8 x 0.5= 0.4 moles of HCL, but for every 2 mol HCl you would make 1 mol product so we would divide 0.4 by 2 to get that the real number of moles is 0.2. For the Zn there's 8.5g/65.37g/mol Zn = 0.13 mol Zn, so there are fewer moles of Zn so it is the limiting reagent.
Then using the enthaplies of formation for the products and reactants you would calculate the $\Delta H_{rxn}^{\circ}$ to be -153.89 kJ/mol. Since only 0.13 mol of Zn is reacting and the $\Delta H_{rxn}^{\circ}$ is for the reaction as written, so 1 mol of Zn, you would multiply $\Delta H_{rxn}^{\circ}$ by 0.13 mol. (-153.89 kJ/mol x 0.13 mol) = -20.0 kJ. This is the heat released by the reaction when the 8.5 g of Zn is added. To find $\Delta T$ you would solve the equation $q=mC_{s}\Delta T$ for $\Delta T$, with q= -20,000 J = (800 g)(4.184 J/goC)$\Delta T$ so $\Delta T$=5.98 so the final T= 31oC.

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