- Oct 28, 2022
yeah i get all thisI am not going to do the calculation for you, but I will outline the method. Water formation has nothing to do with this problem. In this neutralisation reaction, we are donating a proton to ammonia. There is no water formation.
To calculate specific heat capacity, we need three pieces of data:
Q, the amount of heat liberated in the reaction, which we don't know.
m, the total mass of the system, which is 110.5 g
ΔT, the change of temperature of the system, which is 53.7 C
Specific Heat capacity C = Q/(m.ΔT)
I'm pretty sure that the ammonia will be the limiting reagent, but you should check this.
You have to work out how many moles of reaction will occur.
Multiply the moles of reaction by the ΔHreaction to get Q
Do you think you can complete the question now?
would you say i should just treat ammonia as a special case: since basically all other acid base rxn produce water?OK, I understand your question.
It comes back to the different models of acids and bases.
The Arrhenius Model proposes that acids ionise to give H+ ions and bases ionise to give OH- ions. Therefore, under the Arrhenius Model, neutralisation will produce water molecules.
The Bronsted-Lowry Model proposes that acids are proton donors, and bases are proton receivers. Therefore, under the Bronsted-Lowry Model, there is no requirement to produce water molecules during neutralisation. All that has to happen is proton transfer.
this is still mod 6 right?Ammonia is not the only case you will encounter in Chemistry of a Bronsted-Lowry base. Maybe it is time to revise the advantages of the Bronsted-Lowry Theory of Acids and Bases. You are probably going to get a question in the HSC to compare the advantages of Arrhenius and the Bronsted-Lowry theories.
Yes, this is one of the central big ideas of Mod.6this is still mod 6 right?
we haven't gotten up to this yet
Also thank u for all the help!