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Calorimetry Case Essay Sample

Calorimetry Case Pages
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Abstract:
During the experiment, the group were able to perform the following objectives; to compute the heat capacity of a Styrofoam-cup calorimeter, and also to compute the heat of neutralization of 1.0 M hydrochloric acid and 1.0 M sodium hydroxide, the heat of dilution of concentrated sulfuric acid, and the heat of solution of solid ammonium chloride The sixth experiment was named “Calorimetry” wherein it is the measurement of how much heat is gained or released by a system as a chemical reaction occurs within it. The experiment has subtopics namely; Preparation of the Two Styrofoam-Cup Calorimeters, Determination of the heat capacity of calorimeter 1, Heat of neutralization of 1.0 M Hydrochloric Acid and 1.0 M sodium Hydroxide, and lastly Heat of Dilution of Concentrated Sulfuric Acid. The experiment was done in the allotted time and it was quite difficult for us because of the wide range of equation.

Keywords:
Calorimetry, exothermic, endothermic, and heat capacity
Introduction:
Calorimetry is the measurement of how much heat is gained or released by a system as a chemical reaction occurs within it. The heat lost or gained in a chemical reaction is the heat of reaction. The laboratory device in which quantities of heat can be measured is called calorimeter. There are two types of calorimeter: The bomb calorimeter and the open calorimeter. In a bomb calorimeter, the volume of the system is constant. The heat measured equals the change in the internal energy of the system. In an open calorimeter, the pressure of the system is constant.

Heat measurements are made by mixing known amounts of reactants in a calorimeter and letting them react. In an exothermic reaction, the chemical reaction gives off heat to the surroundings. The heat is absorbed by the calorimeter and its temperature rises. In an endothermic reaction, the chemical reaction absorbs heat from the surroundings. The heat is absorbed from the calorimeter and its temperature falls.

The SI unit of energy is the joule (J) and one calorie (cal) is equal to 4.184J. The joule and the calorie are relatively small units for measurements of thermochemical values. The heat capacity (C) of a given mass of a substance is the amount of heat required to raise the temperature of the mass by one degree Celsius. Specific heat (s) is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. Experimental Section:

LIST OF CHEMICALS
1M hydrochloric acid, ammonium chloride, 1M sodium hydroxide, phenolphthalein, Concentrated sulfuric acid, distilled water. LIST OF APPARATUS
2 thermometers, 2 Styrofoam-cup calorimeters, 10-mL graduated cylinder, 50-mL graduated cylinder, iron stand, iron ring, Bunsen burner, wire gauze, electronic balance, 150-mL beaker PROCEDURE
Preparation of the Two Styrofoam-Cup Calorimeters. A Styrofoam-cup calorimeter consists of two Styrofoam cups nested together with a cover. A hole is made in the cover for the insertion of the thermometer. Make sure that the two thermometers will not interchange. Determination of the Heat Capacity of Calorimeter 1. Place 50 mL of cold water (distilled water at room temperature) in calorimeter 1. Put the cover and determine its temperature (Tcold ) by immersing the thermometer for one minute. Place 50 mL of hot water with temperature between 55°C and 60°C in calorimeter 2. Put the cover and determine its temperature ( Thot ) after one minute. Immediately pour the hot water in calorimeter 2 to the cold water in calorimeter 1. Cover and determine the highest temperature ( Tfinal ) reached. Make 2 trials. Calculate the heat lost by the hot water in joules. The specific heat of water is 4.184 J/g-°C. The density of water is 1 g/mL. The mass of water is the product of its volume and density ( qhot = mhots(Tfinal – Thot) ). Compute the heat gained by the cold water in joules. (qcold = mcolds(Tfinal – Tcold)).

Determine the heat gained by calorimeter 1 in joules by applying the law of conservation of energy. (qhot + qcold + qcalorimeter 1 = 0 ,  qcalorimeter 1 = -qhot – qcold ). Calculate the heat capacity of calorimeter in J/°C by dividing the heat gained by calorimeter by the temperature change of the cold water. ( Ccalorimeter = qcalorimeter / (Tfinal – Tcold). Compute the average heat capacity of calorimeter 1 in J/°C. This average heat capacity of calorimeter 1 will be used in other parts of the experiment. Heat of Neutralization of 1.0M Hydrochloric Acid and 1.0M Sodium Hydroxide. Place 50-mL of 1.0M NaOH in calorimeter. Cover and get the temperature (TNaOH) after one minute. Place 50-mL of 1.0M HCl in calorimeter 2. Put the cover and read the temperature (THCl) after one minute. Pour the HCl in calorimeter 2 to the NaOH in calorimeter. Stir using the thermometer and record the highest temperature (Tfinal) reached. Make 2 trials. Calculate the initial temperature using Tinitial = (TNaOH + THCl)/2. Compute the heat gained by the mixture in joules. Assume that the density and the specific heat of the mixture are the same as that of water. The total volume of the mixture is 100-mL. ( qmixture = mmixtures(Tfinal – Tinitial). Determine the heat gained by calorimeter 1 in joules. (qcalorimeter 1 = Ccalorimeter 1(Tfinal – Tinitial) ). Calculate the heat of neutralization in joules by applying the law of conservation of energy. ( qneutralization + qmixture + qcalorimeter 1 = 0, -> qneutralization = -qmixture – qcalorimeter 1 ).

Computer the heat of neutralization per mole of water formed in kJ/mole H2O. Determine the average heat of neutralization per mole of water formed in kJ/mole H2O. Heat of Dilution of Concentrated Sulfuric Acid. Place 97-mL of distilled water in calorimeter 1. Cover and read the temperature (Tinitial) after one minute. Using a 10-mL graduated cylinder, measure 3-mL of concentrated sulfuric acid. Pour the acid in the distilled water in calorimeter 1. Cover and record the highest temperature (Tfinal) reached. Make 2 trails. Calculate the heat gained by the mixture in joules. Assume that the specific heat of the mixture is the same as that of water. The density of the concentrated sulfuric acid is 1.83 g/mL. (qmixture = mmixtures(Tfinal – Tinitial), mmixture = mwater + msulfuric acid). Compute the heat gained by calorimeter 1 in joules. (qcalorimeter 1 = Ccalorimeter 1(Tfinal – Tinitial) ). Determine the heat of dilution in joules by applying the law of conservation of energy. (qdilution + qmixture+ qcalorimeter 1 = 0, -> qdilution = -qmixture –qcalorimeter 1 ). Calculate the heat of dilution per mole of concentrated sulfuric acid in kJ/mole H2SO4. The molar mass of H2SO4 98 g/mole. Compute the average heat of dilution per mole of concentrated sulfuric acid in kJ/ mole H2SO4.

Heat of Solution of Ammonium Chloride. Place 100-mL of distilled water in calorimeter 1. Place the cover and record the temperature (Tinitial) after one minute. Weigh exactly 2.67g of ammonium chloride. Put the ammonium chloride in calorimeter 1. Cover and stir using the thermometer to dissolve the solid. Determine the lowest temperature (Tfinal) reached. Make 2 trials. Calculate the heat lost by the mixture in joules. Assume that the specific heat of the mixture is equal to that of water. ( qmixture = mmixtures(Tfinal – Tinitial) ), mmixture = mwater + mammonium chloride ). Compute the heat lost by calorimeter 1 in joules. ( qcalorimeter 1 = Ccalorimeter 1(Tfinal – Tinitial) ). Determine the heat of solution in joules by applying the law of conservation of energy. ( qsolution + qmixture + qcalorimeter 1 = 0, -> qsolution = -qmixture –qcalorimeter ). Calculate the heat of solution per mole of ammonium chloride in kJ/mole NH4Cl. The molar mass of NH4Cl is 53.45 g/ mole. Compute the average heat of solution per mole of ammonium chloride in kJ/mole NH4Cl.

Results and Discussion:
A. Determination of the Heat Capacity of Calorimeter 1
Trial 1Trial 2
Temperature of Cold Water in °C ( Tcold )30°C30°C
Temperature of Hot Water in °C ( Thot )46°C52°C
Final Temperature in °C ( Tfinal )38°C40°C
Heat Lost by the Hot Water in J ( qhot )-1673.6 J-2510.4 J Heat Gained by the Cold Water in J ( qcold)1673.6 J2092 J
Heat Gained by Calorimeter in J ( qcalorimeter 1 )0 J/°C418.4 J/°C Heat Capacity of Calorimeter 1 in J/°C ( Ccalorimeter 1 )0 J/°C0.024 J/°C Average Heat Capacity of Calorimeter 1 in J/°C0.012 J/°C0.012 J/°C

In this experiment, we were able to determine the heat capacity of calorimeter 1. The table above shows the results of this experiment. Using the equations given in the procedure, we were able to solve the heat capacity of calorimeter. The results positive because this indicates the amount of heat gained or released. But in the end, we got reasonable result as an average. B. Heat of Neutralization of 1.0M Hydrochloric Acid and 1.0M Sodium Hydroxide

Trial 1Trial 2
Temperature of NaOH in °C ( TNaOH )31°C31°C
Temperature of HCl in °C ( THCl )30°C32°C
Temperature of Mixture in °C ( Tfinal )36°C37°C
Initial Temperature in °C ( Tinitial )30.5°C31.5°C
Heat Gained by the mixture in J ( qmixture )2301.2 J2301.2 J Heat Gained by the Calorimeter 1 in J (qcalorimeter 1 )0.066 J0.066 J Heat of Neutralization in J ( qneutralization )-2301.27 J-2301.27 J Moles of water formed0.05 mol0.05 mol

Heat of neutralization per mole of water formed in kJ/mole H2O-46.026 kJ/mol-46.026 kJ/mol Average Heat of neutralization per mole of water formed in kJ/mole H2O-46.026 kJ/mol-46.026 kJ/mol

This experiment is very similar to the first experiment. The difference is just the substance used in the experiment. Computing the heat gained by the mixture, we got 2301.2 J in the first trial and the same amount of heat in the second trial too. The heat of neutralization per mole of water formed we got is -46.026 kJ/mol. The thermal change associated with the reaction of an acid and a base. C. Heat of Dilution of Concentrated Sulfuric Acid

Trial 1Trial 2
Temperature of Water in °C ( Tinitial )31 °C30 °C
Temperature of Mixture in °C (Tfinal )44 °C42 °C
Heat Gained by the Mixture in J ( qmixture )5574.64 J5145.82 J Heat Gained by Calorimeter 1 in J ( qcalorimeter 1 )0.156 J0.144 J Heat of Dilution in J ( qdilution )-5574.796 J-5145.964 J
Moles of Concentrated Sulfuric Acid0.03 mol0.03 mol
Heat of Dilution per Mole of Concentrated Sulfuric Acid in kJ/mole H2SO4-185.83 kJ/mol-171.53 kJ/mol Average Heat of Dilution per Mole of Concentrated Sulfuric Acid in kJ/mole H2SO4-178.69 kJ/mol-178.69 kJ/mol

The thermal change involved when additional solvent is added to a concentrated solution. The results above clearly show what we’ve got by
computation. The initial temperature was 31 °C in the first trial and 30 °C in the second trial. The final temperature were 44 °C and 42 °C. By computation, the heat gained by the mixture is 5574.64 J and 5145.82 J. The heat gained by calorimeter 1 are 0.156 J and 0.144 J. The heat of dilution per mole of Concentrated Sulfuric Acid is -185.83 kJ/mol in the first trial and -171.53 kJ/mol in the second trial. We got both negative in all the trials which means there is a positive heat of dilution in this experiment. D. Heat of Solution of Ammonium Chloride

Trial 1Trial 2
Temperature of Water in °C ( Tinitial )31°C31°C
Temperature of Mixture in °C (Tfinal )28°C28°C
Heat Lost by the Mixture in J ( qmixture )1288.71 J1288.71 J Heat Lost by Calorimeter 1 in J ( qcalorimeter 1 )-0.036 J-0.036 J Heat of Solution in J ( qsolution )-1288.6745 J-1288.6745 J Moles of Ammonium Chloride.04995 moles.04995 moles

Heat of Solution per Mole of Ammonium Chloride in kJ/mole NH4Cl25.8 kJ25.8 kJ Average Heat of Solution per Mole of Ammonium Chloride in kJ/mole NH4Cl25.8 kJ25.8 kJ

In this experiment, we got an initial temperature of 31°C and a final temperature of 28°C. The heat lost by the mixture was 1288.7J. The heat lost by calorimeter 1 is -0.036J. The heat of solution is -1288.6745J. The average heat of solution per mole of Ammonium Chloride is 25.8 kJ/mol. The table results above clearly shows that there is a thermal change that accompanies the dissolving of a substance in a solvent.

Conclusion:
Calorimetry is the measurement of how much heat is gained or released by a system as a chemical reaction occurs within it. The device in which quantities of heat are measured is the calorimeter. There are two types of a calorimeter that is a bomb calorimeter and the open calorimeter. One example of a bomb calorimeter is the one used in this experiment which is the Styrofoam-cup calorimeter it is basically Styrofoam cups with a lid/cover and to measure the heat we need a thermometer. Through the use of the Styrofoam calorimeter we can compute for the heat capacity of the calorimeters which is equal to the beat gained by the calorimeter divided by the temperature of the cold water. The heat of neutralization is the thermal change associated with the reaction of an acid and base and in this experiment we ought to find out what is the heat neutralization of HCL and NaOH and we can determine it by subtracting the negative sign of the heat gained by the mixture by the heat gained by the calorimeter1. The heat of solution of substance is the thermal change that accompanies the dissolving of a substance in a solvent and we can determine it by subtracting the negative of the heat gained by the mixture with the heat gained by the calorimeter1. The heat of dilution is the thermal change involved when additional solvent is added to a concentrated solution and we can determine it by subtracting the negative of the heat gained with the heat gained by the calorimeter1.

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