Dissolution Entropy of Ammonium Chloride Partner Name(s) it is difficult (or imp

Question

Dissolution Entropy of Ammonium Chloride Partner Name(s) it is difficult (or impossible) to to obtain values to plug into We will be using two equations to determine the dissolution entropy because directly measure changes in entropy. The two parts of the experiment allow you the two equations Part 1: Determination of AHsys Raw Data Initial water temperature, Ti 23 2. Final water temperature,T2 20.9°C weight of empaty foam wel vae 019 Weight of ammonium chloride addedo 5. 6. Weight of well vial with water and dissolved NH,C 212 Processed Data (show your work to get credit): Change in temperature of water 8. Moles of solid ammonium chloride dissolved 9. Mass of solution solution (surr) use c 4.18 g °C for the mass, use the entire solution, not just water 0. INH CI (sys) sys

Explanation / Answer

Change of temperature in water:

Final temperature = 20.9 C

initial temperature = 23.3 C

Change in temp = Final temp - Initial Temp = 20.9 - 23.3 = -2.4 C

molar mass of ammonium chloride = 53.5 g/mol

moles = masss/ molar mass

moles = 0.5 / 53.5 = 0.0093458 moles

mass of solution = 17.02 - 16.07 = 0.95 grams

q = mass * heat capacity * temperature difference

q = 0.95 * 4.18 * -2.4 = -9.53 Joules

to find the enthalpy divide the heat by the number of moles

H sys = 9.53 / 0.0093458 = 1019.709 Joules / mole

since the final temperature is lower than the initial one the enthalpy of the system must have positive (+) sign.

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