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# Enthalpy of Neutralisation; with different volumes? watch

1. Hi guys.

25cm3 of 1 mol l-1 sulphuric acid is neutralised by 50cm3 of 1 mol l-1 sodium hydroxide solution. A temperature rise of 9.1 degrees celsius is noted.

Calculate the Enthalpy of Neutralisation.

There is no balanced equation given, simply the info presented there.

My working so far:

4.18x0.075x9.1 = -2.85285

Normally, I would do CxV to give me N then have 1/N x the enthalpy figure.

However, I am unsure here, because of the difference of volumes between the acid and the alkali
2. (Original post by apronedsamurai)
Hi guys.

25cm3 of 1 mol l-1 sulphuric acid is neutralised by 50cm3 of 1 mol l-1 sodium hydroxide solution. A temperature rise of 9.1 degrees celsius is noted.

Calculate the Enthalpy of Neutralisation.

There is no balanced equation given, simply the info presented there.

My working so far:

4.18x0.075x9.1 = -2.85285

Normally, I would do CxV to give me N then have 1/N x the enthalpy figure.

However, I am unsure here, because of the difference of volumes between the acid and the alkali
You need to use the moles of each to determine the limiting reagent.

This then determines the moles of water produced so you can calculate the enthalpy of neutralisation (per mole of water)
3. (Original post by haemo)
moles of limiting (sulphuric acid) = 0.025 x 1
= 0.025

There (d)H = 2.85285/0.025
= -114.0 kJ mol^-1 (exothermic)
Very close, but there is something quite important that you have to realise. Because the ionic equation is H+ +OH- --> H2O you really only want the moles of H+ and OH-. Now, how many moles of hydrogen are there per mole of H2SO4? As a result, what is the concentration of H+ ions?

Think about it before you look at the spoiler!
Spoiler:
Show
There are TWO acidic hydrogens per H2SO4 molecule, so the concentration of H+ is double that of H2SO4 (0.025 x 2). Luckily, this matches the moles of OH- (which is the same as the moles of NaOH) so there is no excess reactant. Therefore, the moles reacted is actually 0.05!

Be careful in future when the moles of H+ and OH- are different. You will need use the one with the fewest moles, since after they have been used up the reaction stops!
4. (Original post by haemo)
I believe:

q = mc(dt)
= 75 x 4.18 x 9.1
= 2852.85 J
= 2.85285 kJ

moles of limiting (sulphuric acid) = 0.025 x 1
= 0.025

There (d)H = 2.85285/0.025
= -114.0 kJ mol^-1 (exothermic)
Your value for the limiting reagent isn't right - think about what sort of acid sulphuric acid is (how many protons it has).

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