### 4.15 7th ed

Posted:

**Tue Feb 12, 2019 2:50 pm**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.

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=76&t=42045

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Posted: **Tue Feb 12, 2019 2:50 pm**

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.

Posted: **Wed Feb 13, 2019 10:47 am**

This question is not as simple as adding a hot piece of metal to a liquid and using q_{liquid}= -q_{metal} 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 q_{liquid}= -q_{metal}. 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:

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 to be -153.89 kJ/mol. Since only 0.13 mol of Zn is reacting and the is for the reaction as written, so 1 mol of Zn, you would multiply 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 you would solve the equation for , with q= -20,000 J = (800 g)(4.184 J/g^{o}C) so =5.98 so the final T= 31^{o}C.

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 to be -153.89 kJ/mol. Since only 0.13 mol of Zn is reacting and the is for the reaction as written, so 1 mol of Zn, you would multiply 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 you would solve the equation for , with q= -20,000 J = (800 g)(4.184 J/g