Thermochemistry

Question

38.5 g of ammonium nitrate was added to 100.0 mL of water. The initial temperature of the water was 25.0 °C. After dissolution, the water temperature was measured at 5.80 °C. What is the enthalpy of solution (energy required to dissolve) ammonium nitrate (J/mol(NH4NO3))?

Answer 1

molar mass NH4NO3 = 80.0 g/mol

q = mC∆T

q = heat = ?

m = mass = 138.5 g assuming a density of water of 1.0 g/ml

C = specific heat of water = 4.184 J/gº (assuming C doesn't change when NH4NO3 is added)

∆T = change in temperature = 5.80º - 25º = -19.2º

Solving for q, we have...

q = (138.5 g )(4.184 J/gº)(-19.2º)

q = - 11,126 J

This is the heat lost (negative sign) when 38.5 g NH4NO3 dissolves. To find the value in J/mol, we need to divide by moles present.

moles present = 38.5 g x 1 mol / 80.0 g/mol = 0.481 mols NH4NO3

∆H = 11,126 J / 0.481 mols = 23,119 J/mol (not corrected for sig. figs.)

Also note that some textbooks/teachers do NOT add the mass of solute (38.5 g) to the mass of water (100.0 g), so you may want to recalculate using 100 g as the value for m in the q = mC∆T equation.