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POSTLAB 9- Heat of Formation of NaCl

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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 in which the exchange of energy occurs during a chemical reaction. Whenever a chemical reaction takes place, heat changes occur with it and these heat changes are measured in the branch of chemistry called thermochemistry. The objective of the experiment is to measure the enthalpy of neutralization for the reaction of the strong base, NaOH, and the strong acid, HCl, and also the heat of formation 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 calorimeter which 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 solution were 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 of the experiment is -52.45 kJ/mol. For the second part, the calculated heat of reaction is 4.09 kJ/mol . With the use of Hess’s law the heat of formation of NaCl (s)  was also derived. The calculated ∆ H of of     NaCl (s) in this experiment is -412.41 kJ/mol.   Therefore, with the use of the experiment, the formula for calculating ∆ H, and Hess’s law , the heats of reactions of NaOH (aq)+ HCl (aq) →  NaCl (aq) +H 2 O (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. The universe 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 an isolated reaction, energy exchange is prevented. To calculate the heat generated per mole of reactant or product, the formula: -m solution c∆ T is used. In this experiment, the heats of reactions of NaOH (aq)+ HCl (aq) →  NaCl (aq) +H 2 O (l),   NaCl (aq) →  NaCl (s),  and the given equations are used to determine the ∆ H of   of NaCl (s) using Hess’s Law . The sum of all equations should be:  Na (s) +  Cl 2(g) →  NaCl (s)   ∆ H of = ?    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  temperatures of individual and combined solutions were recorded using a balance and thermometer.    RESULTS   Table 1. Heats of Reactions in Part A Part A ∆ H Trial 1 -47.7 kJ/mol Trial 2 -57.2 kJ/mol Average Enthalpy -52.45 kJ/mol Table 2. Heats of Reactions in Part B Part A ∆ H Trial 1 4.10 kJ/mol Trial 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)  + ½ O 2(g)  + ½ H 2(g) → NaOH (s) ∆ H= -426.73 kJ/mol (2) NaOH (s) → NaOH (aq) ∆ H= -44.505 kJ/mol (3) ½ H 2(g) + ½ 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 2 O (l) ∆ H= -52.45 kJ/mol (6) NaCl (aq)   → NaCl (s)   ∆ H= - 4.09 kJ/mol  (7) H 2 O (l) → H 2 O (g) ∆ H= +40.668 kJ/mol (8) H 2 O (g) → ½ O 2(g) + H 2(g) ∆ H= +241.84 kJ/mol (9) Na (s) + ½ Cl 2   → NaCl (s) ∆ H= -412.41 kJ/mol *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.    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 in Table 1. Two values were derived for the two trials and the average of those two values was the enthalpy used to calculate for the heat of formation of solid NaCl. This average value is also shown in Table 1. For Part B, solid NaCl was dissolved in water and this was also done in two trials. The results for heat of reaction and the average enthalpy for this experiment is shown in Table  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. The average enthalpy of this reaction will also be used to solve for the heat of formation of solid NaCl. In Table 3, the given set of equations was filled in with the two calculated heats of reactions from the two parts of the experiment to be able to solve for the heat of formation of solid NaCl using Hess’s 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 ∆ H of of solid NaCl which is -412.41 kJ/mol as seen in Table 3. According to the second edition of the book “Principles of General Chemistry”, the ∆ H of of solid NaCl is -411.1 kJ/mol. The deviation of the derived value in the experiment from the theoretical value from the book may be caused by errors made during the experiment. Some sources of error in the experiment can come from errors made by the performer of the experiment or the apparatus. For example, the readings on the volumes of the solutions could be incorrect especially if they were not taken at eye level. Wrong readings with a thermometer can also contribute to errors in measuring the temperature of the solutions. The balance can also give fluctuating readings of the mass and because of this, the experimenter could have recorded the wrong value. Lastly, it is also possible that heat 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 Hess’s law, the heats of reactions of equations (5) and (6) in Table 3, together with the heat of formation of solid NaCl were successfully 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 to change 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 heat generated by the reaction in the coffee cup calorimeter. It is important not to let any energy or heat escape the calorimeter and prevent energy exchange with the surroundings because the heat generated by the reaction will have to be measured  accurately. If the lid is not kept on the container, the measured heat will be wrong. Thus, the calculations for the enthalpies will give erroneous results. 4.   Hess’ s law works by adding the enthalpy changes of the individual steps of an overall process, one will come up with the enthalpy change of that overall  process. Hess’s law is helpful in a way that one doesn’t have to do an experiment and the enthalpy change can still be solved for. We consider the overall reaction as the sum of the series of steps of reactions even if they don’t necessarily occur that way. As long as the ∆ H of the steps are known, it is possible to solve for the ∆ H of the overall reaction. Hess’s law works because ∆ H is a state function, meaning it depends only on initial and final states. 5.   If more concentrated solutions of NaOH and HCl were used in the experiment, the change in temperature will be greater, meaning that the calculated ∆ H will also be greater. On the other hand, the heat of reaction will most likely be the same as the one derived in the experiment because it does not depend on the change in heat or for example, on another factor, the number of moles which will be greater in solutions with higher concentrations. If solid NaOH was used, the heat of reaction will be negatively greater because the dissolution of NaOH (s) is exothermic. The ∆ H of   of NaCl (s)  will most likely also be the same as derived in the experiment, since it is a state function.    SAMPLE CALCULATIONS   Part A Raw Data Table Trial 1 Trial 2 (1) Mass of empty calorimeter + beaker + lid 118.85 g 119.20 g (2) Temperature of Acid (HCl) . C   .  C (3) Temperature of Base (NaOH) .  C .  C (4) Initial Temperature ( ()()  ) .  C .  C (5) Maximum Temperature of Combined Solution .  C .  C (6) Mass of Combined Solutions 268.93 g 270.07 g (7) Mass of Solution Only ( (6)  –   (1) ) 150.08 g 150.87 g Trial 1 ∆ H=   x      m= 150.08 g c= . g C   ∆T= .  C - 28.8   C = .  C  → - .  C because reaction is exothermic Mol= MV = (1M) (0.075 L) = 0.075 mol
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