CH 111 Rutgers University Heat of Neutralization Lab Report Watch my introductory video by Clicking Here. Click and watch each of the following videos. 1 | Course Hero

CH 111 Rutgers University Heat of Neutralization Lab Report Watch my introductory video by Clicking Here.

Click and watch each of the following videos.

1. Experimental Video: Click Here

Document attached below

Using the data provided, perform the calculations need to fill in the missing values on the report sheet. Do not use the data given in the videos.
Submit the lab as an attachment to this assignment (Word document only).

Professor’s Instructions^ CH 111 – Introduction to Inorganic and Organic Chemistry
Lab 8 – Exp. 368 Heat of Neutralization
The reaction between HCl and NaOH generates heat. The goal in this experiment is to find the
change in enthalpy, or heat of neutralization (?Hneut) for this reaction. The heat of
neutralization, ?Hneut, is the amount of heat transferred when 1 mole of an acid reacts with 1
mole of a base.
HCl (aq) + NaOH (aq)
NaCl (aq) + H2O (aq)
The data in this experiment is obtained by monitoring the temperature of the reaction over a
period of 20 minutes. For the first 5 minutes, the temperature of each individual solution,
before mixing, is monitored to find their initial temperatures. Then, at the 5-minute-mark, the
solutions are mixed together, and the temperature is recorded every 30 seconds for the
remaining 15 minutes. This data is then used to generate a graph to determine the final
temperature of the reaction. The method used to determine the final temperature (Tfinal) from
the graph is described in detail in the lab manual. To simplify this, I have included the trendline
and the equation of the line on the graph. Tfinal will be the y-intercept from the equation.
Time (min)
5.5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Mix
28.5
28.5
28.5
28.5
28.4
28.4
28.4
28.4
28.4
28.3
28.3
28.3
28.3
28.3
28.2
28.2
Time vs Temperature – Trial 1
28
27.8
27.6
y = -0.0241x + 27.514
27.4
27.2
27
26.8
0
5
10
15
20
25
Time
Time vs Temperature – Trial 2
29
Tempeerature
Mix
27.4
27.4
27.4
27.3
27.3
27.3
27.2
27.2
27.2
27.1
27.1
27.1
27.1
27.1
27.1
27.1
Temperature
Time (min)
5.5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
y = -0.0207x + 28.628
28.8
28.6
28.4
28.2
28
0
5
10
15
Time
20
25
Supplementary Questions
1. What is the net ionic equation for this experiment?
2. What difference would the use of sulfuric acid, a diprotic acid, have made in your
calculations?
Data Sheet
Trial 1
Trial 2
Volume of HCl solution (mL)
50.0
50.0
Molarity HCl (M)
2.0
2.0
Volume of NaOH solution (mL)
51.0
51.0
Molarity NaOH (M)
2.0
2.0
Initial temp acid (oC)
20.6
20.6
Initial temp base (oC)
21.2
21.2
Moles of HCl (mol)
Moles of NaOH (mol)
Avg Initial temp. of reactants, Tinitial (oC)
Final temperature of mixture, Tfinal (oC)
(from graph)
Temperature change, ?T (oC)
q (heat transferred), (J)
# moles HCl reacting (mol)
?Hneu (kJ/mol)
Mean ?Hneu (kJ/mol)
Calculations
Moles HCl = volume HCl used in Liters (convert!) x Molarity HCl
Moles NaOH req. = volume NaOH used in Liters (convert!) x Molarity NaOH
Final Temperature = From Graph
Temperature Change, ?T = Tfinal – Tinitial
q = specific heat x total mass x ?T
Specific heat of the solution = 3.89 J/g oC
Total mass = (density) x (volume of solution after mixing) (mL)
density = 1.04 g/mL
# moles HCl reacting = Moles HCl (1st calculation)
???
?Hneu ,(kJ/mol) = ?????????? ?????? convert q to kJ, the negative sign is present because we want the
?Hneu for the reaction (the system), not the surroundings. What we “measured” in the
experiment was the surroundings.

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