General Chemistry: When a 4.25 g sample of solid ammonium nitrate dissolves in 60.0 g of water?

Question

in a constant pressure calorimeter, the temperature drops from 22.0 degrees Celsius to 16.4 degrees. Calculate delta H rxn (in kJ/mol NH4NO3) for the solution process listed below.
* Assume that the specific heat of the solution is the same as that of pure water
* NH4NO3(s) --> NH4+ + NO3-

Both answers came out as incorrect.

26.5 kj or -26.5 kj did not come out right.

as for the 31.75 kj i think that is wrong because you molecular wgt. is wrong...

what i mean by wrong is that i turned it in like that and it was wrong.

Answer 1

Water, according to wikipedia, has a specific heat capacity of 4184 Jk^-1kg^-1
Presumably you appreciate that 'per kelvin' is the same as 'per degree celcius'? Just checking.
You have 60.0g of water, which is cooled by 22-16.4 = 5.6 degrees.
How much energy is removed to cool 60 g of water by 5.6 degrees?
4184 J is removed to cool 1000g by 1 degree. Multiply by 5.6 degrees, gives you 23430.4 J.
However, you only have 60 g of water, not 1000g. so, multiply
23430.4 by 60/1000 which gives: 1405.824 J, which is 1.405824 kJ.
You used 4.25 g of ammonium nitrate. How many moles is that? The molar mass of ammonium nitrate is 80.04 gmol^-1.
So, 4.25 g = 4.25/80.04 = 0.053 moles.
To conclude then, delta H is 1.405 kJ per 0.053 moles,
1.405/0.053 = 26.5 kJmol^-1.

Answer 2

Solid Ammonium Nitrate

Answer 3

When a 4.25 g sample of solid ammonium nitrate dissolves in 60.0 g
of water in a coffee-cup calorimeter, the temperature drops from 22.0 *C to 16.9 *C.
Calculate the ΔH for the following reaction:

Answer:

NH4NO3 (s) ---> NH4 (aq) + NO3 (aq)
(assume that the specific heat of the solution is the same as that of pure water
and use your datasheet)
Specific heat of water: 4.180 J/g *C
Amount of heat absorbed: (60.0+4.25 g) X 4.180 J/g *C X (22.0-16.9 *C)
= 1370 J
Molar mass of Amm. Nitrate: 2X14.01+4X1.008+3X16.00= 80.05 g/mol
4.25 g / 80.05 g/mol = 0.0531 mol
So, heat per mol of Amm. Nitrate is 1370 J/ 0.0531 mol = 25800 J
ΔH= + 25.8 kJ/mol {or 26 kJ/mol}
(Endothermic: + H)

Answer 4

calculate the molecular weight of NH4NO3= 2*14+3*16+4 =96g

so 4.25g corresponds to 0.0443mole

the calories lost are dQ = m c dt dt = 16.4-22= -5.6°

m =60g c = 4.186J/cal/°

dQ = -5.6 *60*4.186=-1407J this for 0.0443mole

so -31750J/mole = -31.75 kJ/mole

Answer 5

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