# If the theoretical molar enthalpy of solution for NH4Cl(aq) is +14.7 KJ/mol, what mass of NH4Cl is needed to decrease the temperature of 100.0g of water by -20.0oC.

Write a thermochemical equation for this process, including the energy term.
What would be the molar enthalpy for the reaction used to produce ammonium chloride, as shown below? Use the enthalpies of formation (on the last page) to find this.
NH3(g) + HCl(g)→ NH4Cl(s)

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1. The idea here is that you can use the heat absorbed by the solution to find the heat given off by the dissolution of the salt.

More specifically, you can assume that

Δ H diss=−q
solution

The minus sign is used here because heat lost carries a negative sign.

To find the heat absorbed by the solution, you can use the equation

q=m⋅c⋅ΔT
−−−−−−−−−−−−−

Here
q:
is the heat gained by the water
m:
is the mass of the water
c:
is the specific heat of water
Δ T
is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample
As the problem suggests, you can approximate the mass and the specific heat of the solution to be equal to those of the pure water sample.

The temperature increases by
0.121∘C , so you know that

Δ T= 0.121∘C→
positive because the final temperature is higher than the initial temperature

Plug in your values to find

q=125 g ⋅ 4.18 J g −1∘ C−1 ⋅ 0.121∘Cq
=
63.22 J

So, you know that the solution absorbed
63.22 J
, which implies that the dissolution of the salt gave off
63.22 J
. In other words, you have
Δ H diss = −63.22 J

Convert the mass of sodium hydroxide to moles by using the compound's molar mass

2.4 ⋅10− 4g ⋅1 mole NaOH 39.997g
=
6.00 ⋅10 − 6

moles NaOH

You know that the enthalpy of dissolution when
6.00 ⋅ 10 − 6
moles of sodium hydroxide are dissolved in water, so use this info to find the enthalpy of dissolution when
1
mole of the salt dissolves

1
mole NaOH

63.22 J
6.00

10

6
moles NaOH
=

1.054

10
7

J

Finally, convert this to kilojoules

1.054

10
7
J

1 kJ
10
3
J
=
1.054

10
4

kJ

Therefore, you can say that the enthalpy of dissolution, or molar enthalpy of dissolution, for sodium hydroxide is

Δ
H
diss
=

1.1

10
4
.
kJ mol

1
−−−−−−−−−−−−−−−−−−−−−−−−−−−

The answer is rounded to two sig figs, the number fo sig figs you have for the mass of sodium hydroxide.

SIDE NOTE The accepted value for the enthalpy of dissolution of sodium hydroxide in water at
25

C
is

Δ
H
diss
=

44.51 kJ

Found at: Socratic.org

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2. I just don't get this response by Lola at all. The problem is about NH4Cl, 100 g H2O, and lowering the temperature by 20 C. But the answer is about NaOH, 125 g H2O and raising the temperature by 0.121 C. Add those discrepancies to the struggle at reading the post and taking up 10 pages to post a single page causes even more problems. It appears to me that Lola has attempted to answer a different question.

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DrBob222
3. Here is my response.
How much heat is removed from the water if you want to lower it by 20 C? That's q = mc*dT = 100 g x 4.184 J/g*C x 20 = 8368 J.= 8.37 kJ
How many mols NH4Cl are needed to do that? That's
14.7 kJ/mol x #mols = 8.37 kJ
Solve for mols and multiply by molar mass NH4Cl to convert to grams.
The thermochemical equation is
NH4Cl(s) + H2O --> NH4Cl(aq) delta H = +14.7 kJ/mol
For the last part of the question for this the the following equation,
NH3(g) + HCl(g)→ NH4Cl(s) use this.
dHrxn = (n*dHo formation products) - (n*dHo formation reactants)
Post your work if you get stuck.

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DrBob222