Question

Ammonium nitrate dissociates in water according to the following equation:

43() = 4+()+03−()

When a student mixes 5.00 g of NH4NO3 with 50.0 mL of water in a coffee-cup calorimeter, the temperature of the resultant solution decreases from 22.0 °C to 16.5 °C. Assume the density of water is 1.00 g/ml and the specific heat capacity of the resultant solution is 4.18 J/g·°C.

1) Calculate q for the reaction. You must show your work.

2) Calculate the number of moles of NH4NO3(s) which reacted. You must show your work.

3) Calculate ΔH for the reaction in kJ/mol. You must show your work.

Answer 1

Step-by-step explanation:

NH₄NO₃ = NH₄⁺ +NO₃⁻

heat released by water = msΔ T

m is mass , s is specific heat and ΔT is fall in temperature

= 50 x 4.18 x ( 22 - 16.5 ) ( mass of 50 mL is 50 g )

= 1149.5 J .

This heat will be absorbed by the reaction above .

q for the reaction = + 1149.5 J

2 )

molecular weight of NH₄NO₃ = 80

No of moles reacted = 5/80 = 1 / 16 moles.

3 )

5 g absorbs 1149.5 J

80 g absorbs 1149.5 x 16 J

= 18392 J

= 18.392 kJ.

= + 18.392 kJ

ΔH = 18.392 kJ / mol