Question

So my chem professor had a slide with a question that follows:

If you combine 100 ml of 0.10 M
HCl and 100 ml of 0.10 M NaOH, and find that the temperature goes up 0.67
degrees K  What is ΔH° for the reac/on:
OH(aq)- + H+(aq)->H2O(l)(c H2O= 4.18 J/°C sg, density of water = 1.0 g/ml)

Clicker Q3: (L.G. 5) Solu/on
q =
mcΔT
q = (200 g)(4.18 J/gdeg)(0.67degK)= 560 J
The system is made up of a total of 200 mL of H2O 200 g




THIS is what I don't get because I don't know what the hell L.R stands for.
On my calculator I multiplied the mol of H+ and HO- ( .1 * .1) and it equaled .01.. but I'd like to know if multiplication applies to all the situation of whatever 'L.R' is and if that's the correct way to derive L.R.

Thanks! <3


The enthalpy depends on the number of moles of L.R. 0.01 moles ΔH = q/n = 560 J/0.01 mol = 56000 J/
mol = 56

Answer 1

I expect LR stands for limiting reagent.
Your calculation of 560 J for q is correct.
You added 100 x 0.1M = 10 millimols (0.01 mols) to 0.01 mols
So your answer is 560 J/0.01 mol = 56000 J/mol or 56 kJ/mol

Answer 2

But I thought that the enthalpy of a reaction was dependant on the ratio of moles in the products vs moles in the reactants, not the limiting reagent..