Typical temperature-time graph shape for an endothermic calorimetric experiment

Graphical analysis (e.g. like above) showed the temperature fell by 5.45^oC.

This is obtained using the extrapolated purple lines to correct as much as possible for heat energy gain as the calorimeter and contents start warming up as as the ammonium nitrate dissolves!

 

Calculations

Calculate the enthalpy of dissolution (dissolving) of ammonium nitrate.

NH₄NO₃(s)  +  aq  ===> NH₄⁺ (aq)  +  NO₃^-(aq)

Heat absorbed by the water = mass of water x SHC_water x temperature

q = m c ΔT (m in g, c = 4.18 Jg^-1K^-1, ΔT temperature rise in ^oC)

I'm now going to do the calculation three times, explaining why, but only for example 3.

M_r(NH₄NO₃) = 14 + 4 + 14 + (3 x 16) = 80; mol NH₄NO₃ = 7.2/80 = 0.09 mol

Be aware that different data can be used in the calculation and which is lthe most accurate.

(i) Using the real value of the heat capacity of ammonium nitrate solution for this particular concentration and the real total mass. (The molarity is ~0.9, so SHC solution is ~3.98)

q = m c ΔT

q = 107.2 x 3.98 x 5.45 = 2325 J, 2.325 kJ

Scaling up to 1 mole = 2.325 x (1/0.09) = 25.8 kJ mol^–1

Since the temperature has fallen, the change is endothermic

So ΔH_dissolution(NH₄NO₃) = +25.8 kJ mol^–1   (only accurate to 3 sf)

 

(ii) Using the SHC of pure water with just the mass of the water

q = m c ΔT

q = 100 x 4.18 x 5.45 = 2278 J, 2.278 kJ

Scaling up to 1 mole = 2.278 x (1/0.09) = 25.3 kJ mol^–1

Since the temperature has fallen, the change is endothermic

So ΔH_dissolution(NH₄NO₃) = +25.3 kJ mol^–1   (only accurate to 3 sf)

 

(iii) Using the SHC of pure water and total mass of solution

q = m c ΔT

q = 107.2 x 4.18 x 5.45 = 2442 J, 2.442 kJ

Scaling up to 1 mole = 2.442 x (1/0.09) = 27.1 kJ mol^–1

Since the temperature has fallen, the change is endothermic

So ΔH_dissolution(NH₄NO₃) = +27.1 kJ mol^–1   (only accurate to 3 sf)

 

Comments

Discussion of calculations in terms of accuracy.

(i) This is the most accurate answer because it uses the true specific heat capacity of the salt solution and the actual total mass that is changing temperature.

(ii) This is the next most accurate answer by ignoring the mass of the salt!

It seems that neglecting the mass of the salt roughly off–sets the fall of the heat capacity of water, because all that students will be given will be the higher value of the SHC of pure water.

Some textbooks/internet pages just use the mass of water and its probably the best compromise for the calculation.

(iii) This is the least accurate value calculated.

Although it does use the actual total mass of solution, the real heat capacity of the solution is 5% less. In calculation (ii) the mass is 7% less than (iii) but the SHC of water is 5% higher, so that's why (ii) is more accurate than (iii).