Specific Heat Capacity of Water (Cp,H2O(l)) = 4.184 kJ/kg☌.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.Īlong with the above assumptions, also assume that: This is much closer to what I've seen in the lab when I've done it.Ĭombining all this theory, I made the following graph below which outlines the change in molar heat dissolution and hydrate formation with changing. This effect really kicks in when reaches > 20%wt and results in less temperature rise than expected from calculations when > 20wt%.Īs 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☌ and ending temperature close to 80☌ thus a 70☌ increase making 38%w/w NaOH. Refer below to page 11 of the Dow Caustic Manual: 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.
I've provided good detail below to help anyone else with similar questions. What am I missing?Īfter talking to a very experienced industrial chemist about this he helped me reach a very interesting answer which I believe is correct. The number I calculated is also ridiculously high and obviously wrong. which doesn't make sense as the answer should just be in temperature i.e. 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 via:ĥ59.86 g NaOH x (1 mol NaOH / 39.99 g NaOH) x (-44,500 J / 1 mol NaOH) x (1 mol NaOH (If necessary assume a standard 1L glass beaker with a thermal conductivity = 1 W/m.Molar Heat Capacity of NaOH(s) = 59.5 J/mol.Using that number along with the following thermodynamic properties: The total enthalpy/heat of reaction via Hess' Law is: Assuming the standard enthalpy of formation at 298.15 K / 25☌ (ΔfH) for the following species: I want to predict the maximum and transient temperatures of the chemical system. The dissolution of NaOH is exothermic and the solution gets VERY hot (I physically measured over 70☌+).
As the title suggests, let's say I want to make a 1 L 14M NaOH solution.
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