Thermochemistry

Question

What is the change in temperature when 3.00 grams of ammonium nitrate is dissolved in 100 mL of water? -2.29  ^oC

Given the change in the temperature from adding 3.00 grams of ammonium nitrate to 100.00 mL of water, calculate the enthalpy of solution for ammonium nitrate in units of kJ/mol.

Answer 1

The first question can't be answered unless we know the heat that was absorbed or released. The equation that one would use would be q = mC∆T, where q = heat; m = mass; C = specific heat; ∆T = change in temperature. To find the change in temperature (∆T), we need to know the other variables, which we don't. We only know the mass (100 ml = 100 g) and C = 4.184 J/g/deg. We don't know q, so we cannot solve for ∆T.

But since you give the answer as ∆T = -2.29ºC, we can proceed to answer the second question.

moles ammonium nitrate = 3.00 g x 1 mole/80.0 g = 0.0375 moles

q = mC∆T = (100 g)(4.184)(2.29 deg) = 958 J (this has a positive sign since the temperature decreased makig the dissolution process endothermic)

∆H = 958 J / 0.0375 moles = 25,500 kJ/mol = 25.5 kJ/mol (to 3 significant figures)

(NOTE: depending on the level of chemistry, you may be asked to include the 3 g of ammonium nitrate so the mass, m, would be 103 g instead of 100 g.)