Summary Questions:
1. What value(s) of ΔHrxn did you determine? Compare to the literature value. The first trial that contained 10 mL of NaOH and 10 mL of CH3COOH produced a ΔHrxn of -13.11 kJ/m. The second trial, which contained 15 mL of NaOH and 5 mL of CH3COOH resulted in a ΔHrxn of -7.864 kJ/m. Based on the literature value of NaOH CH3COOH, -57.5 kJ/m, we were off by 44.39 kJ/m in the first trial, and 49.64 kJ/m in the second trial.
2. How did concentration and/or volume differences affect the heat change (q) for each trial? When the volumes of NaOH and CH3COOH were equal, the temperature increased by 5 degrees celsius. When we performed a second trial and added 15 mL of NaOH and 5 mL of CH3COOH, the temperature only showed an increase
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The first part of this lab’s objective was to find the calorimeter constant using DI water. We accomplished this by first checking the temperature and then adding 20 mL of cold DI water into our calorimeter. Next we collected 20 more mL of DI water in a 50 mL beaker and placed it onto a burner in order to heat it. We removed the beaker once the temperature of the water reached 60º celsius. After we removed the water, we poured it into the calorimeter with the cold water and took the temperature. The temperature ended up being 37º celsius. From this information we were able to calculate for qhot, qcold, qcal and Ccal. To be as accurate as possible we conducted this same test three more times and used the averages from all Ccal calculations as the final Ccal. After finding the Ccal of our calorimeter, we were able to integrate the acid and base pair CH3COOH and NaOH. In our first trial we added 10 mL of both CH3COOH and NaOH. Before mixing the two together in our calorimeter, we took the initial temperature. After mixing the two solutions together we took the final temperature. From this point we were able to calculate the qsoln, qcal, qrxn, and ΔHrxn. We conducted a second trial using 5 mL of CH3COOH and 15 mL of NaOH and performed calculations to determine ΔHrxn. …show more content…
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
A higher volume of NaOH will result in more moles of NaOH being added to the HCl, which results in more HCl reacting. This makes the calculated molarity of the HCl be smaller than the actual molarity of the HCl.
The boiling point elevation constant for water that was experimentally determined in this analysis was 0.4396 °C/m, which was derived from the slope of the trend line in Figure 2. This is slightly lower than the constant provided in lecture of 0.51 °C/m. This could be due to further evaporation of water from the solutions tested via refractive index after the boiling temperature was recorded.
The results showed the molarity of the NaOH solution. This experiment was completed twice and a new average molarity
Poured 50 (mL) of sucrose solution from the 250 (mL) Erlenmeyer flask to a 100 (mL) beaker. 14. Warmed the homogenous solution in the 100 (mL) beaker using a hot plate to 38°C. The water needed for each trial was ensured it was heated to 38°C each time using a thermometer. 15. Measured the temperature of the sucrose solution using a thermometer to ensure the temperature was 38°C for equivalency purposes.
For q, you found how much heat was gained by the water so you know that same amount of heat was lost by the metal. Therefore, qmetal = -qwater. The mass of the sample was recorded from the baggie. The temperature change
he experimental variable in this experiment was the changing number of Alka Seltzer tablets that were put into the 250mL beaker of vinegar. Some of the control variables in this experiment were that the same beaker and thermometer were used for each trial, the same type of vinegar and Alka Seltzer were used for each trial, and the same amount of vinegar. The original Hypothesis was that more Alka Seltzer would increase the temperature of the vinegar was not correct. After analyzing the data it can be seen that for the first trial the temperature did not change as it stayed at 24 degrees Celsius before and during the reaction with 1 tablet, also for trial two with two tablets the temperature stayed the same at 23 degrees Celsius and did
The enthalpy also reduces when the temperature of the solution drops, while the concentration of salt and water does not. This results in a smaller energy change and a lower enthalpy since there is less heat and energy. A lower temperature would make the enthalpy even more negative for a process that gives off heat, like NaOH. A lower temperature would result in an enthalpy change that is closer to zero for an endothermic reaction like KCl. However, the enthalpy of the solution would increase if heat entered the cup.
The purpose of this particular lab was to experiment and identify the chemical and physical changes throughout the experiment.
Introduction: The theory behind this experiment is the heat of a reaction (∆E) plus the work (W) done by a reaction is equal to
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.
The control experiment for this investigation will be the experimental setup of 5 trials using 5oC as the temperature. All the steps in the method will be followed.
Purpose: This lab taught procedures for determining heat of capacity of a calorimeter and measuring enthalpy of change for three reactions. It also enforced methods of analyzing data obtained through experimentation and calculating enthalpy. These procedures are used in the branch of thermodynamics known as thermochemistry which is the study of energy changes that accompany chemical reactions. Concepts from this lab can be used to determine the potential energy of a chemical reaction. Much of the energy people depend on comes from chemical reactions. For example, energy can be obtained by burning fuel, metabolizing of food or discharging a batter.
To achieve this, the final value from each thermocouple was set to be equal to the warm water bath temperature (370C), and the initial reading was set equal to the ice water bath temperature. Thus, for each thermocouple an equation was obtained using the two points to convert voltage readings to temperature. An example of the calibration for one of the thermocouples is shown in Appendix II.
This experiment was performed to determine the heat of neutralization between Hydrogen chloride (HCl) and Sodium hydroxide (NaOH). A temperature probe was used to measure the temperature of the reaction when the base (NaOH) was poured into the acid (HCl). The data was collected on logged on LoggerPro.