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
…show more content…
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
If the difference between the recorded times are more than 10%, add a third trial. Repeat these steps with each size beaker. Calculate the average time for each beaker and record the data. The next step of the lab exercise is determining the exact volume of the beakers used in the burn time experiment. Fill a beaker to the top with water. Carefully pour the water from the beaker into a graduated cylinder. Reading the meniscus, record the exact volume into a data sheet. Repeat this step with each size beaker until all volumes are recorded into the data sheet. The final process of the Graphing and Estimating lab is plotting the recorded data onto a graph. Using the data recorded for burn time, in seconds, place the data on the vertical axis. Use the horizontal axis for the volume in milliliters. With the data points plotted determine whether a straight line or a simple curve will best represent the data. Now, obtain a jar and determine the volume. Fill the jar to the top with water. Carefully pour the water into a graduated cylinder. Precisely record the data. Using the plotted data on the graph and the simple curve or straight line to predict how long it would take for the flame to burn out on the candle if it was covered with the jar that was just measured.
13. Calculate the change in temperature for the water caused by the addition of the aluminum by subtracting the initial temperature of the water from the
The dependent variable in the experiment was the temperature and energy absorbed by the water.
Step 3: Use the thermometer to identify and record the temperature for room temperature, in your refrigerator, in your freezer, and then research the temperature of boiling water (do not take this temperature) and record them in Table 2.
This document is not meant to be a substitute for a formal laboratory report. The Lab Report Assistant is simply a summary of the experiment’s questions, diagrams if needed, and data tables that should be addressed in a formal lab report. The intent is to facilitate students’ writing of lab reports by providing this information in an editable file which can be sent to an instructor.
Again with the same temperatures of one degree, eleven degrees, twenty degrees, thirty degrees, and forty degrees. All the temperatures are in Celsius. The one degree C time is at 24.59 seconds, only two seconds off from the first trial. The eleven degrees C time has 20.39 seconds. At twenty degrees C, the data showed the time was 15.54 seconds. Thirty degrees C had the lowest time of the thirty degrees group, at 8.97. The last temperature was forty degrees C with the lowest time at 7.28 (See appendix two). All the data for trial two supports the hypothesis for this
We were also able to conclude that when you combine equal parts of an acid and base you can expect to see a higher temperature change than you would if you added more parts of a base than a acid. Our reasoning behind this conclusion is that when we added 10 mL of both solutions that temperature increased by 5ºc, but when we added 15 mL of NaOH and only 5 mL of CH3COOH there was only a temperature increase of 3ºc. In correlation, we calculated that when you have equal parts of acid and base, the ΔHrxn will be higher than when you use more parts base than
13. The temperature of the water was measured prior to the tube being placed in it and the temperature of the Hydrochloric Acid was measured after it 's temperature had adjusted.
We will be using 6 different fuels to heat up 100ml of water, and find out the changes of the temperature. We will measure the temperatures of the water before and after the experiment. We will burn heat the water for exactly 2 minutes, and check the changes in temperature. The change in temperature will allow us to work out the energy given off the fuel by using this formula:
The lab used methods of calorimetry in order to measure the temperature change of reactions and calculate the changes in
Introduction: The theory behind this experiment is the heat of a reaction (∆E) plus the work (W) done by a reaction is equal to
Use the same Elodea to conduct the other 4 trials for that same temperature, repeat steps 1-5 for each trial.
3. Calculate the total heat released in each reaction, assuming that the specific heat of the solution is the same as for pure water (4.18J/gK). Use q=mcΔT. Show work here and record your answer in Data Table 2.
In order to measure the heats of reactions, add the reactants into the calorimeter and measure the difference between the initial and final temperature. The temperature difference helps us calculate the heat released or absorbed by the reaction. The equation for calorimetry is q=mc(ΔT). ΔT is the temperature change, m is the mass, c is the specific heat capacity of the solution, and q is the heat transfer. Given that the experiment is operated under constant pressure in the lab, the temperature change is due to the enthalpy of the reaction, therefore the heat of the reaction can be calculated.
In this experiment, we investigate the change in temperature caused by adding a chemical substance into the water and dissolving it. The results recorded in the table below show that our hypothesis is correct.