Theoretical temperature change in lime hydration

Calcium oxide (quicklime) and water have an exothermic reaction

$\ce{CaO + H2O \rightarrow Ca(OH)_2} \quad \quad \Delta H^{\circ}_f=-65.3 \; \mathrm{kJ \cdot mol^{-1}}$

Assume that I have 1 kg of lime which means that I would theoretically need 322 grams of water for complete hydration. However, I want to use different amounts of excess water and calculate the final temperature of the whole system after the reaction has finished.

The heats of formation are given at the standard temperature in the tables. Therefore, if I assume that the calcium oxide and water start the reaction at 25 °C I know that 65.3 kJ of heat per mole of reacted $\ce{CaO}$ has to be transferred to the $\ce{CaO}$/$\ce{H2O}$ mixture. With 1 kg of lime (molar mass of $\ce{CaO}$ is 56.08 g) this would mean 1164 kJ of energy released.

So, if for example 1 kg of water was added, the final mixture would consist of 1.322 kg of $\ce{Ca(OH)2}$ and 0.678 kg of $\ce{H2O}$. Using the specific heat capacities the heat capacity of the total system could be calculated and the final temperature rise obtained.

Now, assume that I want to take into account the effect of adding reactants at different temperatures. Say that my slaking water is 16 °C and the lime 35 °C. Should I first calculate the initial temperature by assuming that the reactants do not react until the temperature difference between the lime and the water has settled?

I obtained the heat capacities from here: http://www.update.uu.se/~jolkkonen/pdf/CRC_TD.pdf

$\ce{H2O}$ heat capacity 75.3 J mol$^{-1}$ K$^{-1}$ $\rightarrow$ 4.18 kJ kg$^{-1}$ K$^{-1}$

$\ce{CaO}$ heat capacity 42.0 J mol$^{-1}$ K$^{-1}$ $\rightarrow$ 0,749 kJ kg$^{-1}$ K$^{-1}$

$\ce{Ca(OH)2}$ heat capacity 87.5 J mol$^{-1}$ K$^{-1}$ $\rightarrow$ 1.18 kJ kg$^{-1}$ K$^{-1}$

The amount of heat is calculated from

$Q=cm\Delta t$

If I use the arbitrary initial temperatures of 16 °C for water ($t_1$) and 35 °C for the lime ($t_2$), and initial weights of 1 kg the starting temperature of the system without reaction would then be obtained from

$c_1 m_1 (t-t_1)+c_2m_2(t-t_2)=0$

and solving for final temperature $t$

$t = \frac{c_1m_1t_1+c_2m_2t_2}{c_1m_1+c_2m_2}=18.9 \;°C$

I would then calculate the heat of the reaction, account for the reduction of water and the formation of $\ce{Ca(OH)2}$ and finally calculate the heat distribution and the final temperature.

Is this the proper approach to this problem? How large error is made if the starting temperature of the reaction is not the normal temperature of 25 °C? The temperature range I am interested in would be between 10 °C and 100 °C.

The recommended final temperature for wet slaking is about 71-93 °C. Here I assumed that all of the quicklime reacts and none of the heat is lost to the environment. The reaction rate is dependent on the temperature and in too cold conditions (too much excess water) the hydration can be incomplete as the lime is "drowned". My ultimate goal is to create a model of a slaker and use it to improve the temperature control of a milk of lime plant.

• Are you familiar with Hess's Law? It says that the change in extensive properties (e.g., enthalpy) from the initial state of the system to the final state is independent of path. Reading up on Hess's Law will help you overcome your confusion. Incidentally, 65.3 kJ is not the amount of heat per mole of reacted CaOCaO has to be transferred to the CaOCaO/H2OH2O mixture. It represents the amount of heat that would have to be removed from the reacting system to hold the temperature at 25 C. – Chet Miller Mar 11 '16 at 13:53
• @ChesterMiller I am interested in the final temperature of the system. 1:1 ratio of CaO and H2O might be a bad example as the temperature would rise over 100 degrees as the water evaporates. Nevertheless, if the heat of the reaction is not removed, it has to go to the system itself therefore increasing its temperature. By calculating the heat capacity of the system the theoretical temperature increase could be obtained. To my understanding the reactants have to start from the same temperature so I wanted to ask if my reasoning for the initial temperature was correct. – jannekem Mar 11 '16 at 15:52
• It doesn't matter whether you take that first mixing step or not. The answer will be the same either way. For the adiabatic situation you are interested in, all you need to do is multiply the enthalpy at the final temperature of each of the final reactor constituents by its mass, sum these, and subtract the sum of the enthalpies of each of the initial reactor constituents multiplied by their masses. This should be zero. This will give you an equation for the temperature. – Chet Miller Mar 12 '16 at 0:57
• Alternatively, you can calculate the amount of heat required to heat the initial reactor contents to 25 C, add the heat of reaction times the number of moles converted at 25C, and then add the heat to raise the temperature of the final reactor mixture to the final temperature. This should all sum to zero. – Chet Miller Mar 12 '16 at 1:02
• Ok, now I think i got it. I didn't understand what you meant with multiplying the enthalpies with the masses but I understood the second explanation. So instead of calculating the initial temperature balance I can just calculate the amount of heat needed to be added or removed from the system to bring each reactant to 25C. Then calculate the amount of heat generated, account for the initial addition or removal of heat, and finally use the remaining heat to calculate the final temperature change from 25C. Thanks for your time! – jannekem Mar 12 '16 at 11:44