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

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/g�deg)(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..

Answer 3

The term "L.R" stands for "limiting reactant." In a chemical reaction, the limiting reactant is the one that is completely consumed and determines the amount of product that can be formed.

To determine the limiting reactant in this case, you need to compare the number of moles of HCl and NaOH. You correctly multiplied the molarity of HCl (0.10 M) and NaOH (0.10 M) to find the number of moles of each reactant. Since the reaction is a 1:1 ratio between HCl and NaOH, the reaction will occur until all of one reactant is consumed. In this case, the number of moles of both HCl and NaOH is 0.01 moles.

When calculating ΔH°, you need to consider the number of moles of the limiting reactant. In this case, since the reaction is a 1:1 ratio, both HCl and NaOH will completely react, and the number of moles for each is the same (0.01 mol).

The equation ΔH = q/n calculates the enthalpy change per mole of reactant. In this case, you have determined that the q value (heat absorbed or released by the reaction) is 560 J, and the number of moles of the limiting reactant is 0.01 mol.

By plugging in these values, you can calculate ΔH° as follows:

ΔH° = q/n = 560 J / 0.01 mol = 56000 J/mol = 56 kJ/mol

Therefore, the ΔH° for the reaction is 56 kJ/mol.