How Does Hess's Law Apply to Calculating ΔH in Neutralization Reactions?

[Rapier]

Homework Statement

All solutions are (aq).

In the reaction: HCl + NaOH → NaCl + H2O, ΔH = 60.4 kJ/mol
In the reaction: CH3COOH + NaOH → CH3COONa + H20, ΔH = 51.9 kJ/mol
In the reaction: CH3COOH + NH4OH → NH4CH3COO + H20, ΔH = 51.6 kJ/mol
In the reaction: HCl + NH4OH → NH4Cl + H20, determine ΔH

Homework Equations

Hess's Law: ΔH(reaction) = ƩΔH(products) - ƩΔH(reactants)
ΔH1 = ΔH(NaCl) + ΔH(H2O) - ΔH(HCl) - ΔH(NaOH)
ΔH2 = ΔH(CH3COONa) + ΔH(H2O) - ΔH(CH3COOH) - ΔH(NaOH)
ΔH3 = ΔH(NH4CH3COO) + ΔH(H2O) - ΔH(CH3COOH) - ΔH(NH4OH)
ΔH4 = ΔH(NH4Cl) + ΔH(H2O) - ΔH(HCl) - ΔH(NH4OH)

The Attempt at a Solution

I've never quite understood how this works. All of our previous examples were like 2X + Y = 2B, and had a variation of all three variable in each equation. I think I understand it, and start the problem and 30 seconds later realize I don't understand it at all. I don't see how to make the connection between the first three reactions and the fourth. Please help.

 

[Borek]

Apparently first equation must be taken as it is - that's the only way of having HCl on the left. It also looks like you should add third equation to the first - this way you will have both HCl and NH₄OH on the left. Now you have to think how to use the third equation to get rid of the things you don't need.

Could be writing them all as net ionic will help.

Please remember you can subtract the reaction as well, you don't have to only add them.

 

[Rapier]

Hmmmm. I think I get it...but I've said that before. :)

ΔH1-ΔH2+ΔH3 (Reactants): H + Cl + Na + OH - CH3CO - OH - Na - OH + CH3CO + OH + NH4 + OH
ΔH1-ΔH2+ΔH3 (Reactants): H + Cl + NH4 + OH

ΔH1-ΔH2+ΔH3 (Products): Na + Cl + 2H + O - CH3COO - Na - 2H - O + NH4 + CH3COO + 2N + O
ΔH1-ΔH2+ΔH3 (Products): NH4 + Cl + 2H + O

HCl + NH4OH → NH4Cl + H20

Which proves that ΔH1-ΔH2+ΔH3 = ΔH4, so
ΔH4 = 60.4 kJ/mol - 51.9 kJ/mol + 51.6 kJ/mol
ΔH4 = 60.1 kJ/mol

I looked it up and my results seems to be about 8 kJ/mol higher than it should, but we did this in lab.

 

[Borek]

8 kJ/mol sounds like a reasonable experimental error to me

 

[DeeNos]

Hi there! The concept of Hess's Law can be a bit tricky to understand at first, but it is an important one in thermodynamics. Let's break it down step by step to see how it applies to this problem.

First, let's review the law itself: ΔH(reaction) = ƩΔH(products) - ƩΔH(reactants). This means that the change in enthalpy for a reaction is equal to the sum of the enthalpies of the products minus the sum of the enthalpies of the reactants.

Now, let's look at the first three reactions given:

1) HCl + NaOH → NaCl + H2O, ΔH = 60.4 kJ/mol
2) CH3COOH + NaOH → CH3COONa + H20, ΔH = 51.9 kJ/mol
3) CH3COOH + NH4OH → NH4CH3COO + H20, ΔH = 51.6 kJ/mol

We can see that all three reactions involve the formation of a salt and water. This means we can use these reactions to help us determine the enthalpy of the fourth reaction:

4) HCl + NH4OH → NH4Cl + H20

To do this, we will use the equations you listed, ΔH1, ΔH2, and ΔH3, to calculate the enthalpies of the products and reactants in reaction 4. Let's start with the products:

ΔH(NH4Cl) = ΔH1 = 60.4 kJ/mol
ΔH(H2O) = ΔH2 + ΔH3 = (51.9 + 51.6) kJ/mol = 103.5 kJ/mol

Now, let's look at the reactants:

ΔH(HCl) = ΔH1 = 60.4 kJ/mol
ΔH(NH4OH) = ΔH3 = 51.6 kJ/mol

Plugging these values into the equation for Hess's Law, we get:

ΔH(reaction) = ƩΔH(products) - ƩΔH(reactants)
ΔH(reaction) = (60.4 + 103.5) kJ/mol - (60.4 + 51