Heat of Reaction and Solution Research Assignment Paper (Lab Report Sample)
The main aim of this experiment was to determine the energy between initial and final states of a solution. In this lab, both exothermic and endothermic processes were illustrated. The lab also aimed at measuring the amount of heat absorbed or dissolved in various reactions.source..
Heat of Reaction and Solution
The main aim of this experiment was to determine the energy between initial and final states of a solution. In this lab, both exothermic and endothermic processes were illustrated. The lab also aimed at measuring the amount of heat absorbed or dissolved in various reactions.
Calorimetry is a technique where the heat effect of a given procedure can be measured, with this procedure being a physical or chemical change such as acid-base neutralization. The instrument used is a calorimeter, which can directly measure temperature, thus determining heat effect. Calorimetry is utilized in thermochemistry for the determination of enthalpy and heat capacity among other properties (David 100). Further, energy measurement is a way to analyze the relationship between the properties of a material and energy structure. Enthalpy, H, is given by the following formula:
Where U is the internal energy, P the pressure, and V is the volume of the system, and H is the heat at a constant pressure. Enthalpy changes are used to directly study thermodynamic effects. In this case, a Styrofoam cup calorimeter (calorimetry with constant pressure) was applied. The relations between an amount measured in this calorimeter and the heat effect produced (heat balance equation) are used to give temperature as a function of generated heat from the calorimeter.
In chemical processes, energy differences are mostly observed between the final and initial states due to either heat absorption (endothermic processes) or heat production (exothermic processes) (Hundsdorfer 102). In this experiment, both cases are depicted using the following reactions CITATION Dav10 \l 1033 :
252412551625500HClaq+NaOHaq H2 O+NaClaq(1)
263842543053000HNO3aq+NaOHaq H2 O+NaNO3aq(2)
263842541211500CH3CO2Haq+NaOHaq H2 O+CH3CO2Naaq(3)
252412545021500KNO3s+34H2 Ol KNO334H2 O3(4)
Where (aq) indicates a substance in aqueous solution, (l) a liquid, and (s) a solid. Reactions 1-3 are between acids and bases (neutralizations) with acids in equations 1 and 2 being strong and weak in 3. Further, there is enthalpy of solution, also known as heat of solution in reaction 4 since heat is absorbed at constant pressure (Calvet and Henri 54). In an exothermic process where heat is released, the enthalpy H is negative, while in an endothermic one, the enthalpy is positive. The units of H are given as cal/mole of one of the reactants used or a product consumed. The heat of solution H can be given as:
With the specific heat and amount of KNO3 being small, the equation can be simplified to:
but qreaction= -qcalorimeter
so qcalorimeter= -wwatercwater(tf-t2)
Heat produced in an exothermic process goes to the water in which the products are dissolved. It is noteworthy that the calorimeter determines the warmth of the water. The amount of heat produced is as a result of a given temperature change and can be solved by applying the heat capacity equation below:
Heat change = m C T(units are joules)
Where heat is the product of mass(m), specific heat capacity(C), and temperature change(T), The energy between the initial and final states can, thus, be determined. Also, both exothermic and endothermic processes can be illustrated, and the amount of heat produced in the reactions (1-4) can be measured.
Lab coats, goggles, and gloves were used in this lab to ensure safety.
The insulating calorimeter was a Styrofoam cup with a thermometer and stirring rod. 75ml of 2M NaOH was placed in the cup and the temperature was recorded. 75ml of 2M HCl was added and completely stirred with the thermometer. The highest temperature was noted. The same procedure was repeated for reactions 2 and 3. The temperature change was found in order to calculate the heat absorbed. Further, the number of moles of the reacting acid and base were estimated and H/mole obtained for each neutralization reaction. For reaction 4, 150ml of water (50˚C) was transferred into an 8oz Styrofoam cup. After that, 25g of KNO3 was weighed out with its temperature assumed to be room temperature, ti. The water in the Styrofoam cup was stirred with a stirring bar and its temperature t2 was taken using the thermister. All the KNO3 was added and thoroughly stirred until it dissolved. Finally, the temperature of the system was recorded at intervals of 30 seconds until it was complete tf.
Data and Results
Table 1 shows the data consisting of the amounts of reactants used as in the procedure as well as the calculated results of the heat change and the heat change per mole. The temperature difference in table 1 is obtained using T= Tf -Ti. Calculations for Hrxn and H/mol of acid and H/mol of base are also done and indicated in the table. Further, the mass used was calculated from the product of water density (1.00g/cm3) and the volume of solution in each case. The specific heat capacity was taken to be that of pure water, which is 4.184J/g-deg.
Total volume (ml)
H/mol of acid (J/mol)
H/mol of base (J/mol)
Salt mass (g)
The temperature difference T= Tf -Ti
For reaction 1,
In the same manner, T for reactions 2-4 was obtained as 3.6˚C, 0.3˚C and -4.6˚C respectively.
Calculations for Hrxn and H/mol
Heat change = mCT qrxn = - qcalorimeter Cwater = 4.184 J/gºC
Hrxn= qrxn = mCT = (density of water×volume) × C of water × T
For reaction 1, q= (1×150)×4.184×0.9= 564.84J
However, qrxn = - q, which gives qrxn as -564.84J
In the same way, qrxnfor reactions 2 and 3 were found as -2259.36J and -188.28J respectively.
H/mol of acid (J/mol)
Moles of acid = moles/liter× vol. in liters
For the neutralization reactions 1-3, moles of acid= 2×75/1000= 0.15 mol, thus giving H/moles of acid for reaction 1 as -564.84J/0.15mol= 3765.6J/mol. In the same manner, moles for reactions 2 and 3 were found to be 15062.4J/mol and 1255.2J/mol respectively.
H/mol of base (J/mol)
Moles of base = moles/liter× vol. in liters
For the neutralization reactions (1-3) moles of base= 2×75/1000= 0.15mol
Giving H/moles of base for reaction (1) as -564.84J/0.15mol= 3765.6J/mol and in the same way for reactions (2) and (3) are 15062.4J/mol and 1255.2J/mol.
For reaction 4 of KNO3, qrxn = mCT = (density of water×volume) × C of water × T
Therefore, q= (1×150)×4.184×-4.6= -2886.96J hence qrxn=2886.96...
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