Exothermic: Thermodynamics and Data Table
Heidi Duncan
11/24/13
Exothermic and Endothermic Reactions Lab
The purpose of this lab is to observe how heat is released or absorbed with different chemicals.
Data Table 1 – HCI and NaOH
Trial 1
Trial 2
Avg
Volume 1.0 M HCI(ml)
25
25
-
Volume1.0 M NaOH (ml)
25
25
-
Ti of HCI before mixing
20
20
-
Ti of NaOH before mixing(
20
20
-
Average Ti before mixing(
20
20
-
Tf of mixture )
26
26
-
T )
6
6
-
Specific Heat (J/g)
4.184
4.184
-
Heat, q (J)
1255.2
1255.2
1255.2
Data Table 2- NH4 NO3 and H20
Trial 1
Trial 2
Avg
Mass of NH4 NO3 (g)
12
11.93
-
Volume of H20 ( ml)
25
25
-
Ti of H2O )
20
20
-
Tf of mixture )
-3
-2
-
T )
-23
-22
-
Specific Heat(J/g)
4.184
4.184
-
Heat, q (J)
-2405.8
-2301.2
-
Heat per gram (J/g)
-200
-193
-197
Data Table 3- NaOH and H2O
Trial 1
Trial 2
Avg
Mass of NaOH (g)
7.99
8.06
-
Volume of H2O (ml)
50
50
-
Ti of H20
21
23
-
Tf of mixture
50
55
-
T )
29
32
-
Specific Heat (J/g)
4.184
4.184
-
Heat, q (J)
6066.8
-2301.2
-
Heat per gram (J/g)
759
-831
795
Reaction 1 – HCI and NaOH
1. Determine the average initial temperature for each trial by averaging the initial temperature of the NCI and the NaOH solution before mixing. Record as Average Ti before mixing in Data Table 1.
Substance
Trial 1 (Ti)
Trial 2 (Ti)
Average (Ti)
HCI
20
20
20X 20 /2 = 20
NaOH
20
20
20X 20 /2 = 20
2. Calculate the change in temperature T by substracting the average initial temperature from the final temperature of the mixture (T = Tf –Ti) Record in Data Table 1.
Substance
Tf
Ti
T
HCI & NaOH mix
26
20
26 - 20 = 6
3. Did the temperature rise or fall as the 2 solutions were mixed? Explain this in term of heat transfer.
-The temperature rose as the 2 solutions were mixed because they neutralized.
4. Calculate the heat of neutralization (q in joules) for the reaction using equation (q =m X T X s). Assume that the
11/24/13
Exothermic and Endothermic Reactions Lab
The purpose of this lab is to observe how heat is released or absorbed with different chemicals.
Data Table 1 – HCI and NaOH
Trial 1
Trial 2
Avg
Volume 1.0 M HCI(ml)
25
25
-
Volume1.0 M NaOH (ml)
25
25
-
Ti of HCI before mixing
20
20
-
Ti of NaOH before mixing(
20
20
-
Average Ti before mixing(
20
20
-
Tf of mixture )
26
26
-
T )
6
6
-
Specific Heat (J/g)
4.184
4.184
-
Heat, q (J)
1255.2
1255.2
1255.2
Data Table 2- NH4 NO3 and H20
Trial 1
Trial 2
Avg
Mass of NH4 NO3 (g)
12
11.93
-
Volume of H20 ( ml)
25
25
-
Ti of H2O )
20
20
-
Tf of mixture )
-3
-2
-
T )
-23
-22
-
Specific Heat(J/g)
4.184
4.184
-
Heat, q (J)
-2405.8
-2301.2
-
Heat per gram (J/g)
-200
-193
-197
Data Table 3- NaOH and H2O
Trial 1
Trial 2
Avg
Mass of NaOH (g)
7.99
8.06
-
Volume of H2O (ml)
50
50
-
Ti of H20
21
23
-
Tf of mixture
50
55
-
T )
29
32
-
Specific Heat (J/g)
4.184
4.184
-
Heat, q (J)
6066.8
-2301.2
-
Heat per gram (J/g)
759
-831
795
Reaction 1 – HCI and NaOH
1. Determine the average initial temperature for each trial by averaging the initial temperature of the NCI and the NaOH solution before mixing. Record as Average Ti before mixing in Data Table 1.
Substance
Trial 1 (Ti)
Trial 2 (Ti)
Average (Ti)
HCI
20
20
20X 20 /2 = 20
NaOH
20
20
20X 20 /2 = 20
2. Calculate the change in temperature T by substracting the average initial temperature from the final temperature of the mixture (T = Tf –Ti) Record in Data Table 1.
Substance
Tf
Ti
T
HCI & NaOH mix
26
20
26 - 20 = 6
3. Did the temperature rise or fall as the 2 solutions were mixed? Explain this in term of heat transfer.
-The temperature rose as the 2 solutions were mixed because they neutralized.
4. Calculate the heat of neutralization (q in joules) for the reaction using equation (q =m X T X s). Assume that the