50 mL of copper(II) sulfate reacts with 50 mL of sodium hydroxide. Their concentrations are 0.3 M and 0.6 M respectively. The temperature increased to 23.6 C from 23.4 C.
Determine the enthalpy change for the reaction in kJ/mol of sodium hydroxide.
Q=mcΔT
=(50g + 50g)(4.184 J/g C)(3.2 C)
= 1338.9 J
n=c x v
=0.600 mol/L x 0.050 L
=0.030 mol NaOH
ΔH = -Q/n
= -1.3339 kJ/0.030 mol
= -44.6 kJ/mol
What if it asked for the enthalpy change of the reaction (not specifically just in kJ/mol of sodium hydroxide). When you calculate for moles, would you add the moles of NaOH and CuSO4?
And you know how when you do calculations for the enthalpy change of a reaction you start sort of like this?:
NH3(aq) + HCl(aq) → NH4Cl(aq)
25 mL 25 mL
1.0 mol/L 1.0 mol/L
n = c x v
= 1 mol/L x 0.025L
= 0.025 mol
How come you only need the moles of one of the reactants? Because when we do Q=mcΔT, for m, we include the mass of both solutions... And for example in the first problem I showed, if the volume and concentration of the reactants are different, I don't think it'd be correct to just use the moles of one of the reactants to calculate ΔH for whole reaction...
For other sample problems, the concentrations and volumes of both reactants were conveniently the same. So we never had to choose which reactant to get the moles for (except in the problem I had at the beginning - but that one specifically asks for it in kJ/mol of NaOH). But what if the concentrations and volumes were different and we were never specifically asked to find it in kJ/mol of ,e.g., NaOH? Would that ever happen? Or maybe that's too advanced or something for right now? I'm just scared something like that will pop up on a test and I'm just going to have to guess which reactant to use to find moles...
And it was conveniently the same in this problem, too.
CuSO4 + 2NaOH ==> Cu(OH)2 + Na2SO4
You had 0.3 x 0.05 = 0.015 mols CuSO4.
You had 0.6 x 0.05 = 0.030 mols NaOH but you can see from the stoichiometry that it takes twice as much NaOH as CuSO4. You had only 0.015 mols Cu(OH)2 formed.
Sometimes the reagents DON'T "exactly" match in which case it is just another limiting reagent problem and something will be in excess.
As to why you added 50 mL + 50 mL you need to remember why you do a problem like this. It starts out to find dH for the reaction but you need some method by which to measure it. Therefore, you stick the reaction in a calorimeter and measure the heat produced BY MEASURING HOW MUCH THE WATER IS HEATED. You had 100 mL H2O so you use 100. It makes little difference to the chemicals that 50 came from one and 50 from the other; the idea is that you heated 100 mL H2O, you know the specific heat H2O, and you know the rise in temperature. Therefore, you can calculate the enthalpy change for the water which is (voila) the enthalpy change for the reaction.
When calculating the enthalpy change of a reaction, you only need the moles of one reactant because the balanced chemical equation tells you the stoichiometric ratio between the reactants. In the example you provided, NH3(aq) + HCl(aq) → NH4Cl(aq), the balanced equation shows that 1 mole of NH3 reacts with 1 mole of HCl to produce 1 mole of NH4Cl.
Using the moles of one reactant allows you to determine the amount of heat transferred per mole of that reactant. This is because the balanced equation gives you the stoichiometric ratio between the reactants, indicating the number of moles that react with each other.
In the example where the concentrations and volumes are different between the reactants, you do not need to specifically choose which reactant to use to find the moles. As long as you use the correct stoichiometric ratio from the balanced equation, you will calculate the correct enthalpy change for the reaction. The concentration and volume information is used to calculate the moles of the reactant you choose, but it does not affect the overall enthalpy change.
However, it is always good practice to clearly indicate the reference reactant when reporting the enthalpy change of the reaction. This avoids any confusion and ensures that the enthalpy change is correctly associated with the chosen reactant.
When calculating the enthalpy change of a reaction, you usually focus on one of the reactants or products to determine the moles. This is because the stoichiometric coefficients in the balanced chemical equation give you the ratio of moles between the reactants and products.
In the example you provided, NH3(aq) + HCl(aq) → NH4Cl(aq), we have equal volumes and concentrations of both reactants, which makes it easy to choose one to determine the moles. However, in a more general case where the concentrations and volumes are different, you can choose any of the reactants or products to calculate the moles.
The choice of which compound to use for the moles depends on convenience, where you have the most accurate measurements, or where the numbers are well-rounded for easier calculations. It does not affect the overall enthalpy change of the reaction.
So, in the case where concentrations and volumes are different and you are not specifically asked to find the enthalpy change in terms of a specific compound, you can choose any reactant or product to calculate the moles and use that in the enthalpy change calculation. The final result will represent the enthalpy change of the overall reaction.
In summary, for general enthalpy change calculations, you can choose any reactant or product to determine the moles, as long as the stoichiometric coefficients in the balanced chemical equation are taken into account.