# Exothermic: Thermodynamics and Data Table

Topics: Thermodynamics, Heat, Ammonium nitrate Pages: 8 (777 words) Published: December 12, 2013
﻿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 density and specific heat of the solution are the same as that of pure water (density of water = 1.0 g/ml). Record in data table 1. Calculate the average heat (q) by averaging trials 1 and 2. Record in data table 1. Q=?

M = 50g
S =4.184 J/g
T=26 - 20 = 6
Q=( 50g)(4.184J/g)(6=1255.2J
*Both Trials had the same results
5. Classify the reaction as either exothermic or endothermic. Give evidence for your answer. -The reaction is exothermic because the temperature rose.

Reaction 2 – NH4 NO3 and H20

1. Calculate the change in temperature (T) for each trial. Record in Data Table 2. Trial #
Tf
Ti
T
Trial 1
-3
20
-3 -20= -23
Trial 2
-2
20
-2- 20 -22

2. Did the temperature of the water rise or fall when the NH4NO3 was added. Explain this in terms of heat transfer. -The temperature of the water fell because the NH4NO3 absorbed the heat of the water. 3. Calculate the heat ( q in joules ) for the reaction. Record q for each trial in data table 2. Trial 1

Trial 2
Avg
Q = ?
Q = ?
Q = ?
M=25g
M=25g
M=25g
S=4.184J/g
S=4.184J/g
S=4.184J/g
T= -3 – 20 =-23
T=-2 -20= -22
T=-22.5
Q= (25g)(4.184)(-23)=
-2405.8j
Q=(25g)(4.184)(-22) =
-2301.2J
Q=(25g)(4.184)(-22.5)=
-2353.5
4.Calculate the heat absorbed or released per gram of solute added to the water (in joules/g) record in data table 2. Calculate the average heat per gram by averaging trials 1 and 2. Trial 1
Trail 2
Avg
-2405.8J/12g
-2301.2 J / 11.93 g
-2353.5 J /11.965 g
-200 J/g
-193 J/g
-197 J/g
5.Classify the reaction as either exothermic or endothermic. Give evidence for your answer. -The reaction was endothermic because the temperature fell. 6. What could be some possible uses for a chemical such as ammonium nitrate? -A possible use for ammonium nitrate would be in ice packs. Reaction 3- NaOH and H20

1. Calculcate the change in temperature T for each trial. Record in data table 3. Trial #
Tf
Ti
T
Trial 1
50
21
50=29
Trial 2
55
23
55-23
2. Did the...