POSTLAB 9- Heat of Formation of NaCl

8/10/2019 POSTLAB 9- Heat of Formation of NaCl

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Alreen C. Miranda 20C

9- Heat of Formation of NaCl September 16, 2013

ABSTRACT The universe is divided into two parts called the system and the surroundings inwhich the exchange of energy occurs during a chemical reaction. Whenever a chemicalreaction takes place, heat changes occur with it and these heat changes are measured inthe branch of chemistry called thermochemistry.

The objective of the experiment is to measure the enthalpy of neutralization forthe reaction of the strong base, NaOH, and the strong acid, HCl, and also the heat offormation of NaCl (s).

For the first part of the experiment in which the enthalpy of the neutralization of NaOH and HCl is unknown, the acid and base were mixed in a coffee cup calorimeterwhich was used to trap the heat from leaking out from the system and into the

surroundings before taking note of the temperature changes and the mass of the solution.In the second part of the experiment, temperature changes and the mass of the solutionwere also recorded when solid NaCl was dissolved in water.

Using the formula to determine H, the value for the heat of reaction in Part A ofthe experiment is -52.45 kJ/mol. For the second part, the calculated heat of reaction is4.09 kJ/mol . With the use of Hesss law the heat of formation of NaCl (s) was also derived.The calculated Hof of NaCl (s) in this experiment is -412.41 kJ/mol.

Therefore, with the use of the experiment, the formula for calculating H, andHesss law , the heats of reactions of NaOH (aq)+ HCl (aq) NaCl (aq)+H 2O (l) and

NaCl (aq) NaCl (s), and the heat of formation of NaCl (s) were derived.

INTRODUCTION

Thermochemistry measures the heat changes that accompany chemical reactions. Theuniverse is divided into two: the system, the part to be studied, and the rest, the surroundings.A system can release (exothermic) or absorb (endothermic) heat to its surroundings. In anisolated reaction, energy exchange is prevented. To calculate the heat generated per mole ofreactant or product, the formula: -m solution cT is used.

In this experiment, the heats of reactions of NaOH (aq)+ HCl (aq) NaCl (aq) +H 2O (l), NaCl (aq) NaCl (s), and the given equations are used to determine the H

of of NaCl (s) using

Hesss Law . The sum of all equations should be:

Na (s)+ Cl2(g) NaCl (s) Hof = ?

METHODS

First, 75 mL of 1 M HCl and 75 mL of 1 M NaOH were mixed in a coffee cup calorimeter. Next, 4.38 g of NaCl (s) was dissolved in 150 mL water. Initial and final masses and


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temperatures of individual and combined solutions were recorded using a balance andthermometer.

RESULTS

Table 1. Heats of Reactions in Part APart A HTrial 1 -47.7 kJ/molTrial 2 -57.2 kJ/mol

Average Enthalpy -52.45 kJ/mol

Table 2. Heats of Reactions in Part BPart A HTrial 1 4.10 kJ/molTrial 2 4.07 kJ/mol

Average Enthalpy 4.09 kJ/mol

Table 3. Given Set of Equations and the Heat of Formation of NaCl (s) (1) Na (s) + O 2(g) + H 2(g) NaOH (s) H= -426.73 kJ/mol(2) NaOH (s) NaOH (aq) H= -44.505 kJ/mol(3) H 2(g) + Cl 2(g) HCl (g) H= -92.30 kJ/mol(4) HCl (g) HCl (aq) H= -74.843 kJ/mol(5) NaOH (aq) + HCl (aq) NaCl (aq) + H 2O (l) H= -52.45 kJ/mol(6) NaCl (aq) NaCl (s) H= - 4.09 kJ/mol (7) H 2O(l) H 2O(g) H= +40.668 kJ/mol(8) H 2O (g) O 2(g) + H 2(g) H= +241.84 kJ/mol(9) Na (s) + Cl 2 NaCl (s) H= -412.41 kJ/mol

*The value of H for equation (6), NaCl (aq) NaCl (s), became negative because thereverse reaction from Part B is used so that it will be possible for the substances to becancelled.

DISCUSSION In Part A of the experiment, HCl and NaOH were mixed and the initial and final masses

and temperatures were recorded. Using the formula H= , the heat of the reaction

was determined. This was done in two trials and the results of the calculations are seen inTable 1. Two values were derived for the two trials and the average of those two valueswas the enthalpy used to calculate for the heat of formation of solid NaCl. This averagevalue is also shown in Table 1.

For Part B, solid NaCl was dissolved in water and this was also done in two trials. Theresults for heat of reaction and the average enthalpy for this experiment is shown in Table


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2. It is important to mention that only the enthalpies were averaged and not the raw data.These values were arrived at using the same formula in Part A of the experiment. Theaverage enthalpy of this reaction will also be used to solve for the heat of formation ofsolid NaCl.

In Table 3, the given set of equations was filled in with the two calculated heats ofreactions from the two parts of the experiment to be able to solve for the heat offormation of solid NaCl using Hesss law. The value of H for equation (6), NaCl (aq)

NaCl (s), became negative because the reverse reaction from Part B is used so that it will be possible for the substances to be cancelled.

The reactions and their H were added to come up with the Hof of solid NaCl which is-412.41 kJ/mol as seen in Table 3. According to the second edition of the bookPrinciples of General Chemistry, the Hof of solid NaCl is -411.1 kJ/mol. The

deviation of the derived value in the experiment from the theoretical value from the bookmay be caused by errors made during the experiment.

Some sources of error in the experiment can come from errors made by the performer ofthe experiment or the apparatus. For example, the readings on the volumes of thesolutions could be incorrect especially if they were not taken at eye level. Wrongreadings with a thermometer can also contribute to errors in measuring the temperature ofthe solutions. The balance can also give fluctuating readings of the mass and because ofthis, the experimenter could have recorded the wrong value. Lastly, it is also possible thatheat escaped the coffee cup calorimeter and the measured heat will give erroneous results

when used in the calculation of the enthalpies. (AtQ 1)

With the use of the formula H= and Hesss law, the heats of reactions of

equations (5) and (6) in Table 3, together with the heat of formation of solid NaCl weresuccessfully determined.

ANSWERS TO QUESTIONS 1. (Answered in Discussion)2. Heat capacity is the amount of heat needed or required to change the temperature

of an object by 1 K while specific heat capacity is the amount of heat needed tochange the temperature of 1 gram of an object or a substance by 1 K.

3. Keeping the lid on the container is important to trap and keep all the heatgenerated by the reaction in the coffee cup calorimeter. It is important not to letany energy or heat escape the calorimeter and prevent energy exchange with thesurroundings because the heat generated by the reaction will have to be measured


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H=( )( )( )

x

= -47.7 kJ/mol

Trial 2

H= x

m= 150.87 gc= . g C T= . C - 28.2 C= 6. C -6. C because reaction is exothermicMol= MV = (1M) (0.075 L) = 0.075 mol

H=( )( )( )

x

= -57.2 kJ/mol

Average Enthalpy=

= -52.45 kJ/mol

Part BRaw Data Table

Trial 1 Trial 2(1)Empty calorimeter mass 119.29 g 119.37 g(2) Mass of water (150 mL) 266.28 g 265.27 g(3) Mass of water only ( (2)-(1) ) 146.99 g 145.90 g(4) Water temperature . C . C

(5) Lowest Temperature afterDissolution . C . C

Trial 1

H= x

m= 146.99 gc= . g C T= . C - . C= - . C . C because reaction is endothermic

Mol= 4.38 g NaCl x

= 0.075 mol

H=( )( )( )

x

= 4.10 kJ/mol

Trial 2

H= x


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m= 145.90 gc= . g C T= . C - . C = - . C . C because reaction is endothermic

Mol= 4.38 g NaCl x

= 0.075 mol

H=( )( )( )

x

= 4.07 kJ/mol

Average Enthalpy=

= 4.09 kJ/mol

Solving for the Heat of Formation of NaCl (s) Using Hesss Law (1) Na (s) + O 2(g) + H 2(g) NaOH (s) H= -426.73 kJ/mol(2) NaOH (s) NaOH (aq) H= -44.505 kJ/mol(3) H 2(g) + Cl 2(g) HCl (g) H= -92.30 kJ/mol(4) HCl (g) HCl (aq) H= -74.843 kJ/mol(5) NaOH (aq) + HCl (aq) NaCl (aq) + H 2O (l) H= -52.45 kJ/mol(6) NaCl (aq) NaCl (s) H= - 4.09 kJ/mol (7) H 2O(l) H 2O(g) H= +40.668 kJ/mol(8) H 2O (g) O 2(g) + H 2(g) H= +241.84 kJ/mol(9) Na (s) + Cl 2 NaCl (s) H= -412.41 kJ/mol

Na (s) + O 2(g) + H 2(g) + NaOH (s) + H 2(g) + Cl 2(g) + HCl (g) + NaOH (aq) + HCl (aq) +

NaCl (aq) + H 2O(l) + H 2O (g) NaOH (s) + NaOH (aq) + HCl (g) + HCl (aq) + NaCl (aq) + H 2O (l) +NaCl (s) + H 2O (g) + O 2(g) + H 2(g) =

Na(s) + Cl2 NaCl(s) H of = -412.41 kJ/mol

*The value of H for equation (6), NaCl (aq) NaCl (s), became negative because thereverse reaction from Part B is used so that it will be possible for the substances to becancelled.

CONLUSION The heats of reactions of the equations NaOH (aq)+ HCl (aq) NaCl (aq)+H 2O(l) and

NaCl (aq) NaCl (s) were determined using the formula H= . The heat of reaction for

NaOH (aq)+ HCl (aq) NaCl (aq)+H 2O (l) is -52.45 kJ/mol and 4.09 kJ/mol for NaCl (aq) NaCl (s). With the use of Hesss law, the heat of formation of NaCl (S) is -412.41kJ/mol.


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REFERENCES Gross R, Abenojar E, Tan J. Modern Experiments in General Chemistry I. 9 th ed. QuezonCity: Department of Chemistry, Ateneo de Manila University; 2011.

Silberberg M. Principles of General Chemistry: Second Edition. 2010. McGraw-Hill.

POSTLAB 9- Heat of Formation of NaCl

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