# How do I calculate the maximum temperature of an exothermic NaOH reaction in water?

As the title suggests, let's say I want to make a 1 L 14M NaOH solution. For this I'd need:

• (14 mol/ L) x 1 L x (39.99 g / mol ) = 559.86 g NaOH(s) {solute}
• ~1 L H2O (approx) {solvent}

The dissolution of NaOH is exothermic and the solution gets VERY hot (I physically measured over 70°C+). I want to predict the maximum and transient temperatures of the chemical system. Assuming the standard enthalpy of formation at 298.15 K / 25°C (ΔfH) for the following species:

• ΔfH_NaOH(s) = -425.6 kJ/mol
• ΔfH_Na+(aq) = -240.1 kJ/mol
• ΔfH_OH-(aq) = -230 kJ/mol

The total enthalpy/heat of reaction via Hess' Law is:

-(-425.6 kJ/mol) + (-240.1 kJ/mol) + (-230 kJ/mol) = -44.5 kJ/mol

Using that number along with the following thermodynamic properties:

• Molar Heat Capacity of NaOH(s) = 59.5 J/mol·K
• (If necessary assume a standard 1L glass beaker with a thermal conductivity = 1 W/m·K)
• Initial ambient temperature of 298 K.

How can I determine the max temperature that the solution will reach? I've tried doing calculations myself with these constants but I get answers with units [mol·K] via:

EDIT 1:

559.86 g NaOH x (1 mol NaOH / 39.99 g NaOH) x (-44,500 J / 1 mol NaOH) x (1 mol NaOH·K / 59.5 J) = -10,471 mol NaOH·K = -10744.15 mol NaOH·°C

...which doesn't make sense as the answer should just be in temperature i.e. [K]. The number I calculated is also ridiculously high and obviously wrong. What am I missing?

• How do you get a number in mol×K? Need to see the calculations. Commented Feb 17, 2022 at 9:47
• @OscarLanzi refer to the newest edit cheers. I know the answer will likely involve some sort of integration for the transient/dynamic temperature but I'm not really sure how... Commented Feb 17, 2022 at 13:52
• The heat capacity of water is different from the heat capacity of the NaOH solution. So the same amount of Joules will produce a different temperature increase in the beginning and at the end of the dissolution. So you would have to know the formula describing how the heat capacity of the solution changes with the concentration. This is not an easy task. Commented Feb 17, 2022 at 18:02
• As someone with bachelors in both chemistry and chemical engineering, I can tell you that this sort calculation was not focused on in my chemistry courses, but is an important and standard exercise in multiple chemical engineering courses. If you search for "Adiabatic temperature rise", you can find calculation methodologies at varying levels of approximation. Commented Apr 15, 2022 at 20:38
• This is very useful! If you did more work on the industrial 48-50% to 25% operation I'd like to discuss with you or read your final report. On the plant we use a fin fan to cool it after mixing, but winter can be challenging in terms of freezing of caustic since this is also an interesting non linear curve and should be considered for design of applications. Commented Jul 28, 2023 at 8:23

After talking to a very experienced industrial chemist about this he helped me reach a very interesting answer which I believe is correct. I've provided good detail below to help anyone else with similar questions.

1. The Theory/Logic:

The first pointer he mentioned is that the molar heat of solution of NaOH in water does not hold true at NaOH concentrations approaching 14M NaOH, this is because as NaOH (s) dissolves it can form NaOH.nH2O as well as Na+ and OH- with (n = 1,2,,3.5,4...7 (etc.)) all depending on the concentration of NaOH. Refer below to page 11 of the Dow Caustic Manual:

This effect really kicks in when [NaOH] reaches > 20%wt and results in less temperature rise than expected from calculations when [NaOH] > 20wt%. As an example experiment form literature validating this, refer to the graphic below from "Experimental Study of the Thermal Effect of the Dissolution Reaction for some Alkalis and Salts with Natural Mixing an Forced Stirring" below showing starting temperatures of Starting temperature 10°C and ending temperature close to 80°C thus a 70°C increase making 38%w/w NaOH. This is much closer to what I've seen in the lab when I've done it.

Combining all this theory, I made the following graph below which outlines the change in molar heat dissolution and hydrate formation with changing [NaOH]. It can be seen that the actual molar heat of dissolution I should be using is not -44.5 kJ/mol but really around -30 kJ/mol.

2. The Calculation/Application:

Along with the above assumptions, also assume that:

• Specific Heat Capacity of Water (Cp,H2O(l)) = 4.184 kJ/kg°C
• Solute Mass = (14 mol/ L) x 1 L x (39.99 g / mol ) = 559.86 g NaOH(s)
• Assume 14M NaOH density is 1460 g/L
• Solvent Mass = (1460 - 559.86)g = 900.14 g (i.e. NaOH is 38.3wt%)

Dissolving 559.86g NaOH (for 14 M) in 1L of water produces an enthalpy change (∆H) equivalent to -30kJ/mol x 14mol = - 420 kJ.

Taking the equation ∆H = mCp∆T and rearranging to solve for ∆T = ∆H/(m x Cp) = -420 kJ / (1.46 kg x 4.184 kJ/kg) = 68°C increase Therefore, Tf = Ti + ∆T = (24.85 +68)°C = 93°C

This value is much closer to the value obtained in the experiment graphic above as well as what I originally measured in the lab.

I would suspect that if I made the 14M NaOH solution in a 1L pyrex beaker with open top, losses from the conduction of heat into beaker, radiation from beaker walls and conduction loses from liquid surface would keep the liquid closer to 70-80°C.

If the 14M was made in a larger quantity e.g. 100L in a 200L capacity drum, I would expect that the true 68°C increase in temperature over ambient to be realized as losses would be small relative to the amount of heat generated. Also, I acknowledge that a more accurate result would come about by measuring the actual density of 14M NaOH and using in calculations.

Many plastics are not capable of holding liquids at 93°C especially if their density is 1.5kg/L so, as another inferential answer to my question, using a metal container is certainly preferable.

EDIT1 & UPDATE:

Just thought I'd add the extrapolated curves I got based on these calculations by changing concentration. The graph below includes a tail at the end after a smoothing of data series, I believe this is from the 6th degree polynomial I used to extrapolate the heat of dissolution trend where R^2 =/= 1. So for concentrations below, say 19, with this method seem more accurate. Assumes an overall 95% heat transfer efficiency.

Adding more practicality to this (which is what I'm aiming for with my research), is the graph below used with industrial "50% caustic" (my product's spec actually had a 49.9% alkalinity) which can be diluted with water to achieve various concentrations. Here instead I'm assuming 100% efficiency.

Hope this makes sense and is useful to others as it has been for me!

• Why do you insist on dissolving in 1 L water when you just determined the water mass 900 g? Another mistake is considering the solution has the specific heat capacity of water while significant part of it is not water. // Note that the molar ratio=f(conc %) can be formulated as the exact function. // See meta.stackexchange.com: new feature table support for easy syntax to format the text into a table, which is preferred to screen captures or scanned images. Commented Mar 14, 2022 at 7:59
• I suggest using some inert object (e.g. from stainless steel or glass) and measure its heat capacity in a water bath with different initial water and object temperatures. Then measure the heat capacity and calculate the specific heat capacity of the target 14 M NaOH by doing the same with this solution. // Be aware the higher volumes mean higher peak temperature due slower heat dissipation due lower surface/mass ratio. Commented Mar 14, 2022 at 8:22
• I am sorry your "answer" from March 14 is not easy to understand. For example, your first table gives values of $\Delta H$ for the dissolution of 1 mole NaOH in $n$ mol water. But this $\Delta H$ values is useless it the heat capacity of the solution is not given. And this heat capacity is changing during the dissolution process. Why not give the maximum temperature change instead of the $\Delta H$ values ? Second you do not explain the difference between natural mixing and forced stirring. Natural mixing is probably done at the bottom of a beaker. But the result will not be homogeneous. Commented Mar 21, 2022 at 17:22
• Note that the first table and figure are from the Dow Caustic manual. It's from their experiments I've just used some of their results to build correlations. Commented Dec 2, 2022 at 8:59