To install StudyMoose App tap and then “Add to Home Screen”
Save to my list
Remove from my list
In the past, I have experienced numerous injuries while playing sports. At times, my coach used to give me an instant cold or hot pack to provide me some relief from the pain. I never fully understood how by cracking the bag, it would immediately change the temperature of the bag. Eventually, I found out that this is related with a change in heat, and during that moment a chemical reaction takes place. In a cold pack, ammonium nitrate (NH4NO3) is dissolved into water to decrease the temperature of the surroundings (Risko, 2019).
In contrast, a hot pack dissolves calcium chloride (CaCl2) or magnesium sulfate (MgSO4) into water to increase the temperature of the surroundings (Risko, 2019).
This sparked my curiosity, as I began to ask myself several questions: Do all reactions have this effect? Would a greater presence of a substance lead to a larger change in temperature? I then decided to look at neutralization reactions to see if there was any change in the temperature of the surroundings.
Verified writer
4.8 (309)
“ Writer-marian did a very good job with my paper, she got straight to the point, she made it clear and organized ”
I chose to look at the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH), as hydrochloric acid can be found in the human stomach acid. Hence, I wanted to see how heat may be released or absorbed in our body (Koch, 2016).
Furthermore, I decided to have five different volumes (cm3) of hydrochloric acid to answer my earlier question – would the amount of a substance present affect the temperature change. To ensure the experiment can be conducted in a school laboratory, I decided to have 4.00 g of sodium hydroxide in each trial.
For my calorimeter, I chose to use a polystyrene cup as I already knew that it is a great heat insulator. Once understanding the different components of the experiment, I questioned myself if the reactions would affect the molar enthalpy (kJ mol-1) of the reaction (ΔH). This led to the development of my research question: How does the change in volume (cm3) of hydrochloric acid (HCl) in a neutralization reaction with 4.00 grams of sodium hydroxide (NaOH) affect the molar enthalpy (kJ mol-1) of the reaction (ΔH)?
NaOH (s) + HCl (aq) → NaCl (aq) + H2O (l)
Enthalpy (ΔH) is the difference in the internal energy of a closed system with the addition of volume and pressure, however this investigation will keep both variables constant (Atkins, 2011). Specifically, molar enthalpy is the enthalpy calculated on a per mole basis for the substance. In thermochemistry, there are often two types of reactions: endothermic and exothermic reactions. An exothermic reaction occurs when the enthalpy is a negative value, therefore has lost energy in the system and increased the temperature of the surroundings (Bylikin, 2014). In contrast, an endothermic reaction is when the enthalpy value is positive, increasing the energy in the system and decreasing the temperature surrounding it (Bylikin, 2014).
This investigation will focus on a neutralization reaction between hydrochloric acid and sodium hydroxide. A neutralization reaction occurs between an acid and base to form salt and water. An acid is a substance that produces hydrogen ions, H+ (Bylikin, 2014). An alkali base produces hydroxide ions, OH-(Bylikin, 2014). The combination of both substances causes one substance to neutralize the other. The heat change of this reaction is known as the enthalpy of neutralisation. Majority of neutralisation reactions are exothermic, as bonds are created, therefore energy is being released. Furthermore, both hydrochloric acid and sodium hydroxide are strong acids and bases, which would result in a higher enthalpy value as they would both dissociate completely during the reaction.
There are many factors that affect the enthalpy of a reaction. Temperature, concentration, volume and the state of the substances all influence enthalpy. This investigation will explore the effect of volume on the molar enthalpy of a neutralisation reaction.
It is hypothesized that as the volume of hydrochloric acid increases, so does the enthalpy, however the molar enthalpy will remain the same. With greater volume, there are more particles and bonds that can be formed. Hence, this would mean more energy being released, causing an exothermic reaction. The molar enthalpy will remain the same as the enthalpy is calculated by a per mole basis. Therefore, when it is used in the equation, q = mcΔT, then divided by the number of moles, both values will cancel each other out, keeping the final answer constant.
Independent Variable: Volume of hydrochloric acid, measured in centimetres cubed (cm3). This is the independent variable because this is being changed throughout the experiment and is not dependent on any other value. The volumes used were of hydrochloric acid were: 5.00 cm3, 10.0 cm3, 15.0 cm3, 20.0 cm3 and 25.0 cm3. These values were chosen as they provide a wide range of measurable results. Small values were chosen because if larger volumes were used, it would cause the limiting reagent to become excess, therefore providing inaccurate results. There would not be enough sodium hydroxide to react with the acid, making the base the limiting reagent.
Dependent Variable: Molar enthalpy of the reaction, measured in kilojoules per mole (kJ mol-1). This is the dependent variable because it is reliant on the volume of the limiting reagent, hydrochloric acid.
Molarity of hydrochloric acid: The molarity of HCl used in the experiment was kept constant at 1.00 M. A constant molarity ensures accurate results to make sure that only the volume was being evaluated, and not concentration. To achieve this concentration, 31.25 cm3 (31.3 cm3) of hydrochloric acid was diluted in 343.75 cm3 (344 cm3) of water (calculations are shown later). The molarity was kept low, as hydrochloric acid is classified as a very strong Arrhenius acid.
Mass of sodium hydroxide: The mass of NaOH used in the experiment was kept constant to 4.00 grams to ensure that only the limiting reagent was altered, which was the acid. Changing both reactants proportionally would result in the same 1:1 ratio, therefore having no effect on the enthalpy change. Since NaOH is a very strong base, only 4 grams of it was used. Hence, the reaction would be safe and easily conductible.
Pressure: The pressure of the reactants used in the experiment was kept constant, as it would alter the results of the experiment. As stated by Hall in 2015, the equation of enthalpy is H = E + pV, p representing pressure (Hall, 2015). Changing the pressure would affect the enthalpy, therefore pressure was kept constant at the average atmospheric pressure of 1013.25 millibars (1.01 x 103 millibars) (mb) or 101.3 kilopascals (1.01 x 102 kilopascals) (kPa) (Goldstein, 2002). There may be small fluctuations in pressure, however the experiment was conducted at one altitude in the same room, therefore the pressure remained constant.
Time for completion: The time for the completion of the reaction was kept constant at 3 minutes. This time was chosen, as it would provide enough time for the reaction to fully use up its reactants. All the reactions were completed at around 2 min and 45 seconds, as this is indicated by the maximum heat released from the reaction.
Prepare a solution for HCl with a molarity of 1.00 M by diluting 31.25 cm3 (31.3 cm2) of HCl with 343.75 cm3 (344 cm3) of tap water using a 1.00 x 102 cm3 graduated cylinder (±0.5 cm3) to measure and pour into the 5.00 x 102 cm3 volumetric flask. Ensure that safety gloves, goggles and full sleeves are worn to avoid skin contact. Insert and secure a rubber stopper at the opening of the volumetric flask. Shake the flask for 3 minutes to ensure a complete mixture.
Using two 2.00 x 102 cm3 polystyrene cups, cut one cup halfway horizontally using scissors. Once the cut is completed safely, poke a small hole at the bottom of the half cup to ensure there is enough space for the temperature probe to be inserted. Obtain a small sized plastic weigh boat and place it on the electronic balance (±0.01 g). Zero the scale, and from the bottle weigh 4.00 grams of NaOH. Once the scale reads this value, remove it from the electronic balance. Using the diluted HCl solution, pour 25.0 cm3 from the 5.00 x 102 cm3 volumetric flask into the 1.00 x 102 cm3 graduated cylinder (±0.5 cm3).
From the 1.00 x 102 cm3 graduated cylinder (±0.5 cm3), pour the solution into the full-sized 2.00 x 102 cm3 polystyrene cup. Insert a Vernier temperature probe, connected to a Vernier Lab Quest Pro Data Logger (±0.1 °C), into the solution. Start the stopwatch and wait for one minute. Record this temperature into a data table, then insert the 4.00 g of NaOH into the solution in the full 2.00 x 102 cm3 polystyrene cup, and immediately cover the cup with the half cup previously cut out.
Start stopwatch and record the temperature in 30 second intervals into a data table for 3 minutes.
Once 3 minutes has passed, remove the temperature probe and the top cup. Dispose of this solution into a waste container.
Repeat steps 4-9 for four additional trials to obtain accurate results. Repeat steps 4-9 for each volume: 5.00 cm3, 10.0 cm3, 15.0 cm3 and 20.0 cm3 (±0.5 cm3) of HCl.
Safety Considerations: HCl (aq) is a very strong acid, that could severely impact an individual. The National Institute for Occupational Safety and Health (NIOSH) summarizes that HCl is very corrosive and can easily burn the skin or eyes (NIOSH, 2018). Furthermore, if the substance is inhaled or swallowed, it can poison the individual. To avoid this, safety goggles, full sleeves and safety gloves were worn to avoid any contact with the skin.
Ethical and Environmental Considerations: Since the chemicals used are harmful to the school’s sewage system, the solutions were disposed in a waste container, to avoid drainage in the sink. Ensure that there are no tripping hazards around the working station to avoid potential injury.
When the NaOH was added to the HCl solution in the 2.00 x 102 mL polystyrene cup, a popping sound could be heard. This was because of the reaction that was occurring. The cup itself began to slightly heat up, as my hand was able to feel the heat of it. I noticed steam (H2O (g)) coming out from the edges of the cup, suggesting that the heat was escaping. This is significant as it shows how heat generally moves upwards, to a cooler region in this case. Once the top half cup was taken off after the reaction, the solution appeared to be very white. The NaOH had completely dissolved into the solution, therefore it became a homogenous solution from a heterogenous solution originally composed of HCl (aq) and NaOH (s).
Pre-lab calculation for the volume of hydrochloric acid and water needed to create a 1.00 M for the acid
c_1 V_1=c_2 V_2
V_1=((1.00)(.375))/12.0 V_1=0.03125 (dm)^3 or 31.3 (cm)^3
C2 was the desired concentration, which was 1.00 M. V2 was the volume required for the experiment, which was calculated by multiplying 5 by each trial volume and adding the values together. C1 was the molarity of the original HCl. The water required for the solution would be: 375 cm3- 31.25 cm3 = 343.75 cm3 (343 cm3). This was a pre-lab calculation that needed to be completed to determine the volumes required.
Pre-lab sample calculation to determine which reactant is the limiting reagent for 25.0 cm3 of HCl and 4.00 g of NaOH
c= n/V n= m/M
n=(1.0 M)(0.025 〖dm〗^3) n= 4/((22.99+16.00+1.01))
n=0.0250 mol HCl n=0.100 mol NaOH
To calculate the number of moles for HCl, the concentration equation was used, since HCl had a molarity of 1.00 M. To calculate the number of moles for NaOH, the molar mass equation was used, since it was in its solid state. As provided by the calculations, the number of moles in the highest volume of HCl is lower than the number of moles in 4.00 grams of NaOH, therefore HCl is the limiting reagent. Mole ratio does not need to be considered, as it is a 1:1 mole ratio.
Sample calculation of determining the initial and final temperature for 25.0 cm3 of HCl and 4.00 g of NaOH
T_i= (23.7+20.2+20.6+20.5+22.2)/5 T_f= (68.7+64.1+64.7+63.7+41.7)/5
T_i=21.4 °C T_f=60.6 °C
To calculate the initial and final temperature, the mean for each was calculated using the 5 trials that were conducted.
Sample calculation for determining the heat transferred to the surroundings for 25.0 cm3 of HCl and 4.00 g of NaOH
q=mc∆T
q=(25.0 g)(4.18 J g^(-1) K^(-1))(60.6 ℃-21.4 ℃)
q=4096.4 J or 4.10 kJ
To calculate the heat that was lost to the surroundings, the above formula was used. The specific heat capacity was of water, since the solution had been diluted in water. The mass used was specifically only the mass of HCl. In this equation, we assume that the density of HCl is 1.00 g/cm3.
Sample calculation for determining the molar enthalpy for 25.0 cm3 of HCl and 4.00 g of NaOH
Molar Enthalpy= (-q)/n
∆H=(-4.0964 J)/(0.025 mol HCl)
∆H=-163.856 kJ/mol or -164 kJ/mol
In this equation, q (heat) is a negative value, since heat is only measured through the surroundings. Molar enthalpy (ΔH) is the difference in the internal energy of a closed system per mole, therefore the heat gained by surroundings will be negative in the system (Atkins, 2011). The number of moles was calculated earlier in the report. Since the molar enthalpy value is negative, this means that it is an exothermic reaction, as the products have lost energy, making it more stable than the reactants (see figure 1 earlier in the report).
Example calculation of the average percentage uncertainty for the temperature values at 25.0 cm3 of HCl and 4.00 g of NaOH
((0.1/45*100%)+(0.1/43.9*100%)+(0.1/44.1*100%)+(0.1/43.2*100%)+(0.1/19.5*100%))/5=0.284%
To calculate the average percentage uncertainty for the Vernier Lab Quest Pro Data Logger, the absolute uncertainty was divided by the measured values of temperature at 25.0 mL of HCl.
Sample calculation of the average percentage uncertainty for the mass of 4.00 g of NaOH
0.01/4*100%=0.25%
The mass was the same for each trial, therefore the measured value was 4.00 g. Since it was a digital measuring device, the absolute uncertainty is the lowest significant figure.
Sample calculation of the average percentage uncertainty for measuring the volume at 25.0 cm3 of HCl
0.5/25*100%=2.00%
There was only one volume for this example that needed to be measured using a graduated cylinder.
Sample calculation of the total percentage uncertainty at 25.0 cm3 of HCl
0.284%+0.25%+2%=2.53%
This value is a sum of all the percentage uncertainties for the devices used to find the molar enthalpy of 25.0 cm3.
Calculation to show the random error for the entire experiment
(2.53%+3.03%+3.82%+5.54%+10.6%)/5=5.10%
This value represents the random error of the entire investigation. See Appendix A Table 7 for the completed table of all the calculated values.
After the analysis provided above, the molar enthalpy of the neutralization reaction remained constant, however the heat released continued to increase. The results support my hypothesis because as the volume of HCl increases, so does the enthalpy, however molar enthalpy will remain constant. Using the q=mc∆T equation, when the mass increases, so does the difference in temperature, ultimately leading to a larger value. However, to calculate molar enthalpy this value must be divided by the number of moles. Since both the numerator and the denominator increases proportionally, both values cancel out resulting in the same answer for each trial conducted.
According to Mitra, a professor at Missouri State University, the change in internal energy is the sum of the energy that enters and leaves the system (Lucas, 2015). Mathematically this can be represented by the following equation: ∆U=Q-W. W represents work and under constant pressure, the system must “work” to meet the volume change, leading to the equation, Q = ∆U+P∆V. An increase in volume will increase the enthalpy. For this neutralization reaction specifically, it is exothermic as there are many bonds being formed. Bond formation is an exothermic process, therefore causing ∆H to be a negative value. When bonds are broken, energy is required to break it, therefore being an endothermic process.
Furthermore, since the heat gained by the surroundings was a positive value, this meant that the reaction lost energy, resulting in an exothermic process. Bond formation is an exothermic process because of the molecular orbital theory. It states that when a bond is formed between two atoms, the electrons move from a higher energy orbital to a lower one, and the energy is released into the surroundings (Halpern, 2019).
Supported in the data above, there is a clear trend that the molar enthalpy generally stays constant. It peaked at times due to other potential errors or outliers discovered during the experiment. In the graph, most of the values for molar enthalpy remained constant. In Table 6 of the Appendix, it clearly shows how the amount of heat gained by the surroundings was increasing as volume increased. Some of the values throughout the experiment do not match the general trend, due to potential errors, which was shown through the random error of the investigation.
For this experiment, accuracy could not be looked at as there are no other studies that have conducted this experiment at 1.00 M of HCl. However, the precision of the experiment varied. The final temperatures of the reactions achieved varied from 0.00 °C to 2.00 °C, without mentioning outliers. The precision can be expressed by the random error of the experiment, which was 5.10%. This high value means that the precision is low, and this can be due to the experimental instruments used during the experiment.
Investigating HCl Volume's Impact on Neutralization Enthalpy. (2024, Feb 19). Retrieved from https://studymoose.com/document/investigating-hcl-volume-s-impact-on-neutralization-enthalpy
👋 Hi! I’m your smart assistant Amy!
Don’t know where to start? Type your requirements and I’ll connect you to an academic expert within 3 minutes.
get help with your assignment
