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Calculating Standard Heats of a Reaction

Essay by   •  May 8, 2018  •  Lab Report  •  2,306 Words (10 Pages)  •  237 Views

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Lab 5: Calculating Standard [a]Heats of a Reaction

Abstract:[b]

The heat of reaction lab was designed for students to calculate the standard enthalpy of a system. Enthalpy is the study of heat liberated and absorbed from a physical and chemical view. Enthalpy in other words is the flow of heat in a system. A system is defined as the products and reactants. While the surround is the things that isn’t apart of the system. The method that was used in this experiment to figure out the heat of reaction was using the calorimeter to figure out the heat change with the heat capacity that was given. The results for the first part of the heat of a solution per mole of KNO3 (s) dissolved was 42.6014 kJ/mol with a 22.1 percent error. The reaction type was endothermic. The results for the second part in heat neutralization per mole of H30+ was -56.18kJ/mol with a 1.5 percent error. The reaction for the second part was in fact exothermic. The results of the last part heat of reaction per mole of Mg (s) reacted was -458.88 kJ/mol with a 0.78 percent error. The results of the reaction was exothermic which means temperature of the reaction increased and with such low percent error it means that the reaction was correctly calculated.

Introduction: [c]

The change in enthalpy is what is considered the heat of a reaction. The change in enthalpy is important to this lab because it teaches the students concepts of whether a reaction is endothermic or exothermic. Endothermic can be defined as the absorption of heat while exothermic is the liberation of heat. In chemistry it is known that energy can not be created nor destroy so there for this lab is important because it shows how temperature not only changes a substance from a chemical state but also a physical one as well. Another concept that was learned in this experiment is heat capacity. “Heat capacity is the specific is the amount of heat required to raise the temperature of 1 gram of the substance by 1°C”(Thermochemistry,1) Heat capacity is used to help indicate how much is needed to raise a substance temperature. The heat capacity for the lab was 4.184 J/ °C. It was constant throughout all the parts in the lab.

In this lab the only way the enthalpy of a system can be calculated is by a calorimeter. The way that a calorimeter was achieved with the simple equipment in the lab is that it was set up with two coffee cups (polystyrene). The calorimeter was set up using two cups, a stabilized a fiber board to cover the cups and a thermometer that used to acquire the temperature of the substance.  Enthalpy was calculated due to the “standard conditions of temperature and constant pressure of the reactants”( Lucas pg.1) 1. The overall purpose of this lab is calculating the enthalpy of a system using the calorimeter. The interesting thing about this experiment is that the first law of thermal dynamics can be seen clearly through the reaction. The first law of thermodynamic says that nothing can be lost or gained in a system. However, we can see the changes through the exchanges of endothermic and exothermic reaction processes. Thermodynamics is defined as the branch of physics that explains the relationship between heat and other forms of energy.

This information is critical to chemistry because it helps the students understand how exactly molecules react with thermal energy.  

Method:[d]

This experiment was divided into three parts

Part 1:

 Mass  of the polystyrene cup, stir bar, and the 100 mL of deionized water was measured using an analytical balance. Then the mass of potassium nitrate was measured using an aluminum weighing dish. After that the temperature of the water before adding potassium nitrate was measured. Then the final temperature of the potassium nitrate once it became fully dissolved was measured. To get the change in temperature the final temperature and the initial must be subtracted. When the change in temperature has been acquired then heat capacity can be calculated. Heat capacity can be found by using the mass of the substance then multiplying it by 4.184J/(Celsius * grams) and then multiplying that by the change in temperature.

Part 2:

 Polystyrene cups and the stir bar were washed with deionized water. Reuse them and add 50 mL of 1.00  M hydrochloric acid solution. In another graduated cylinder measure 30 mL to 40 mL of 1.00 M sodium hydroxide solution. The temperature was measured with the thermometer of the hydrochloric acid solution in the cup before you add sodium hydroxide solution and after you add the sodium hydroxide solution. The heat capacity can be calculated, by using the mass then multiplying it by 4.184J/(Celsius * grams) and then multiplying that by the change in temperature of the reaction.

Part 3:

The polystyrene cup, stir bar and thermometer probe were washed with deionized water like part 2.75 mL of 1.00M hydrochloric acid solution was measured using a measuring cylinder into the polystyrene cup. After that determine the mass of the cup, stir bar, and hydrochloric acid.  With sandpaper scratch up a fifteen-centimeter strip of magnesium ribbon then weigh it. Then get the temperature of the hydrochloric acid solution in the polystyrene cup before and after adding the Magnesium ribbon strip. Make sure to constantly stir the mixture for one minute until it reached either the maximum or the minimum final temperature. After that make sure to dump the waste in the correct bins and clean your area.

 Data and Calculations: [e]

1.

                             

Table 1: The Data of the Heat of Potassium Nitrate

Table 1: This table shows the data that was acquired through measurements and afterward

Using the measurements to get the heats of the solution.

a. Mass of empty cut with stir bar

8.5761g

b. Mass of cup, stir bar and 1000mL

105.27g

c. Mass of water

96.7g

d. mass of weighing dish

0.92g

e. Mass of weighing dish plus KNO3(s)

2.9816g

f. Mass of KNO3

2.0619g

g. Mass of solution

98.8622

h. Heat capacity of Calorimeter

413.639J/Celsius

i. Initial temperature

23 Celsius

j. Final temperature

20.9 Celsius

k. Temperature change

-2.1 Celsius

l. Calorimeter heat change  

-868.6428J

m. solution heat change

+868.6428J

n. moles of KNO3

0.02039mol

o. Heat of a solution per mole of KNO3 (s) dissolved

42.6014 kJ/mol

p. Is dissolving KNO3 in water exothermic or endothermic

Endothermic

q. Calculate the percent error in your result

heat of solution +34.89 kJ/mol

22.1%

...

...

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