i need d - g answered and below i have the information. if not legible, it says:
ID: 971504 • Letter: I
Question
i need d - g answered and below i have the information.
if not legible, it says: a student determined the delta H neutralization for the reaction of sodium hydroxide solution and acetic acid, using the procedure described in this module. the student added 100.0ml of 0.8500M NaOH to 100.0ml of 0.8404M HC2H3O2. prior to mixing, the temperatures of the base and acid were 17.68°C; the maximum temperature of the mixture was 23.25°C.
(d) Cakulate the heat gained by the calorimeter cup, temperature probe, and stirring bat, l Assume that the calonmeter constant was 29.1 J/0. Teat (e) Determine the total heat evolved (k) in the neutralization reactiont (0) Cakculate the number of moles of acetic acid neutralized by the NMOH modes acetic acid (8) Calculate the molar heat of neutralization of acetic acid, AHoExplanation / Answer
Initial temperature = 17.68
Final temperature = 23.25
a) Rise in temperature = 23.25 - 17.68 = 5.57
b) mass of solution = Volume X density = (100 + 100) X 1.02 = 204 grams
c) Heat gained = Mass X specific heat X change in temperature = 204 X specific heat X change in temperature
Specific heat is not clear, it should be near 4.18
Heat gained = 204 X 4.18 X 5.57 = 4749.65 Joules
d) The calorimeter constant = 29.1
Rise in temperature = 5.57
so heat absrobed by calorimeter cup etc = 29.1 x 5.57 J = 162.087 Joules
e) Total heat evolved = Total heat absorbed = 4749.65 Joules+ 162.087 Joules = 4911.737 Joules = 4.911 KJ
f)
The moles of NaOH added = molarity X volume in litres = 0.85 X 0.1 = 0.085
The moles of acid = = molarity X volume in litres = 0.8404 x 0.1 = 0.08404 moles
So moles of acid neutralized = 0.08404 moles
g) Molar heat of neutralization = Total heat evolved / moles neutralized = 4.911 KJ / 0.08404 moles= 58.44 KJ / mole
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