Enthalpy Changes

[deep0199]

It would be greatly appreciated if anyone can tell me how to find out the enthalpy change for the dissolution of ammonium chloride?

[Dak]

its +314.55 Kj/mol. [1]

 

as for how you work it out, its been a while since iv done any chemical equasions but i believe youd work out how many mols of NH4Cl you have, measure the temperature of the water, ditch the ammonium chloride in the water and measure the temperature change.

 

translate this temperature change from K to Kj, and divide by the number of mols[NH4Cl] to get Kj/mol. now flip the sighn, and thats your {delta}Hf [NH4Cl]

 

i think...

[deep0199]

Can anyone confirm this please?

[jdurg]

That is correct. The simple procedure is as follows:

 

1): Have a known volume of water and determine the mass of the water either via weight, or from the density and the volume data.

 

2): Weigh a known quantity of ammonium chloride.

 

3): Have a good thermometer or thermocouple to measure the starting temperature of the water.

 

4): Make sure everything is done in a thermally isolated container. (I.E. use a styrofoam cup of Dewar flask to make sure there is no energy loss/gain from the surroundings).

 

5): Dissolve the known mass of the ammonium chloride in the thermally isolated water.

 

6): Measure the resulting temperature when it reaches a steady point.

 

7): Using the change in temperature, the mass of water and the specific heat of water, calculate how many joules were exchanged in the dissolution.

 

8): Divide this amount by the number of moles of ammonium chloride you used.

 

9): If the change in temperature was a positive change (I.E. the water was warmer after dissolution), then the sign on your answer will be negative. If the water was cooler after dissolution, then the sign on your answer will be positive.

 

That's it.

[Guest lilangel13]

Can anyone help me in finding a way to compare the acidity of alcohols, phenols and carboxylic acids without using indicators?

[budullewraagh]

check the dissociation constants

[Guest lilangel13]

well i mean practically but thanks anyway