There was a lot of error in the experiment but the two trials are the correct nu
ID: 509385 • Letter: T
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There was a lot of error in the experiment but the two trials are the correct numbers, via answered by another chegg expert.
As for the heat capacity of the thermocup, we're assumingis is 0, (which is obviously incorrect), but that's one of the assumptions that are listed in the procedure of the lab. Thanks for your help!
Think of the neutralization of acetic acid as taking place in two steps: Remember that HCI = H_3O and that NaOH = OH^-. This reaction is the same as the one you calculated in # 2 above. a. Calculate the average value for the molar heat of neutralization for your two trials. This gives you the delta H degree for reaction (3). (You have already calculated delta H degree for equation (2) using HCl and NaOH above.) b. Use Hess' law and your experimental results from the two neutralization reactions to calculate the delta H degree for reaction (1), the dissociation of a weak acid. This is an example of how a delta H that is difficult to measure directly can be determined from other more easily measured quantities. Experiment details: Polystyrene drinking cups are placed inside of each other to serve as the calorimeter. Inner cup will be covered with a lid to minimize heat gain/loss during a chemical reaction Heats of neutralization of a week acid: Know this a big question but I am beyond lost. Please help me it would be very appreciated.Explanation / Answer
Ans. #1. Calculation of heat of neutralization: Trial 1:
Moles of acetic acid used = Molarity x Volume in liters
= (0.5 M) x 0.050 L
= 0.025 mol
Moles of NaOH used = Molarity x Volume in liters
= (0.9817 M) x 0.040 L
= 0.039 mol
In a balanced stoichiometry reaction, 1 mol acetic acid is neutralized by 1 mol NaOH.
In the given reaction mixture, acetic acid is the limiting reactant.
# Total mass of the reaction mixture = 50 mL acetic acid + 40 mL NaOH
= 90 mL
= 90 g ; [assuming density to be 1.0 g/ mL]
# Heat produced during neutralization = - heat gained by calorimeter system
= - m c dT
= - (90 g) x (4.184 J g-10C-1) x 50C
= - 1882.8 J
# Since the reaction mixture contains only 0.025 mol acetic acid, the heat released is solely by neutralization of 0.025 moles acetic acid.
Now, molar enthalpy of neutralization =
heat released per mol acetic acid
= (- 1882.8 J) / 0.025 mol
= - 75312 J/ mol
= - 75.312 kJ/ mol
Ans. #2. Calculation of heat of neutralization: Trial 2:
Heat produced during neutralization = - heat gained by calorimeter system
= - m c dT
= - (90 g) x (4.184 J g-10C-1) x 60C
= - 2259.36 J
# Since the reaction mixture contains only 0.025 mol acetic acid, the heat released is solely by neutralization of 0.025 moles acetic acid.
Now, molar enthalpy of neutralization =
heat released per mol acetic acid
= (- 2259.36 J) / 0.025 mol
= - 90374.4 J/ mol
= - 90.37 kJ/ mol
Note: A. Heat gained by water is given by-
q = m x s x dT - equation 1
Where,
m = mass of water in gram
s = specific heat of water = 4.184 J g-10C-1
dT = change in temperature = T1- T1
B. The -ve value of enthalpy indicates that heat is being released during neutralization of acid.
C. Please recheck your experimental values and strength of acid/ base to get a value closer to the theoretical value of molar enthalpy of neutralization.
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