- Processing The Data You Make Your Own Repot Sheet For This Experiment 1 Prepare A Table With The Following Headings R 1 (245.02 KiB) Viewed 300 times
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PROCESSING THE DATA You make your own repot sheet for this experiment. 1. Prepare a table with the following headings: Reaction, Mass of NaOH, Initial Temperature, Final Temperature, Heat Energy Released, Moles of NaOH and Molar Heat of Reaction. Next, create a row for each reaction. Complete the table using your collected information and any needed calculations. 2. Write the net ionic equations for Reactions 1, 2 and 3. Consider the heat given off in your reactions, add variables to your equations to represent heat evolved in kJ/mole. Use X, for Reaction 1, X2 for Reaction 2. and x3 for Reaction 3 (as written above). 3. The energy. X-, in Reaction 1 represents the energy of solution for one mole of NaOH(a). Look at the net ionic equations for Reactions 2 and 3. and make a similar statement concerning the significance of x2 and X3. 4. Find the difference between the value of X, and the sum of X, plus X3. Account for any similarity or difference. 5. Calculate the percent difference between X, and the sum of X, and X3. (Assume X2 to be correct.) 6. Suppose you had used 4.00 grams of NaOH, in Reaction 1. What would have been the number of Joules released in the reaction? What effect would this have on the value of x.? DISCUSSION 1. Provide an explanation of the results collected above 2. Look up Hess' Law and discuss how this experiment is an illustration,
Data Reactions Mass of NaOH Initial Temperature Final Temperature NaOH, - Na 2) + OH an Heat Energy Released? 1109.3 J Moles of NaOH? 0.025 Mole Molar Hest of Reaction? 44.4 KJ/mol 25.0°C 30.30 °C 25.0 °C NaOH fao H2O + Na + C1 37.0 °C 2511.6J 0.025 Mole 100.5 KJ/mol 25.0° C 31.7 °C 1402.3 J 0.025 mole 56.09 KJ mol Naag) + OH a Cl 4 -f!O@ +Na° {a@ C1-29 + xkJ
v Reaction I ✓ Reaction2 ✓ Reaction 3 Mass H20=50g. C H2O = 4.186 J/g.c = final temp - initial temp =30_30– 25 = 5.3 deg C Mass of HCI= 50 ml = 50g final temp - initial temp = 37 - 25 = 12° C Volume of NaOH=25 mL of 1.0 M = 25 x 1/1000 mole = 0.025 mole Total volume = 50 mL = 50 g dt = final temp - initial temp=31.7-25= 6.7 °C = mass Q=mass* c* At = 50*4.186 12 = 2511 6 J Q=mass* c* At 50g*4.186 Jg.c*5.3°C = 1109.3 J = $0 x 4.186 x 6.7 J = 1402,3 J Mole of NaOH=mM =ig 40mol = 0.025 Mole Mole of NaOH =mM = 1g 40mol = 0.025 Mole AH= 1402.3 0.025 J AH=25116.1/0.025 mol = 100464 J/mol = 100.5 KJ/mol AH= 1109.3/0.025 = 44372 J/mol = 44.4 KJ/mol = 56092 I'mol = 56,09 KJ/mol I
2. Net ionic Reaction 1 NaOH(s) > OH(aq) +Na++X1 KJ Reaction 2 NaOH(s) + H+ (aq) → Na+ (aq) + H20 (1) + X2 KJ Reaction 3 H* + OH → H2O (1) + X3 KJ Xi= 44.4 KJ, X2= 100.5KJ and X3= 56.09KJ 3. Xl represents the energy of solution for one mole of NaOH solid. X2 is the energy of solution for one mole of NaOH solid in water and neutralization of base with HCI X3 is is the energy of neutralization reaction between NaOH and HCI. X2 - X1=X3 ] = 100.2 - { 14.4+ 56.09] 100.2 - 100.49 = -0.29 the difference is almost equal to zero. Thus X2 = X1 + X3 % difference = [0.29 / 100.2] x 100 = 0.29% 6 If we used 4 g of NaOH in reaction 1 the no of joules released would be 4 times what is released in the calculation 4 x 1109.3 J = 4437.2 J
Moles of NaOH = 4/40=01 Therefore X1 = 44272 / 0.1 = 44372 J = 44.4 KJ This will have no effect of the molar heat of solution because it is fixed per mole of compound Discussion 1 This experiment taught important notions. After this lab, I understood how a colorimetry works, to measure the heat of a solution and put Hess's law in practice. I learned that energy or heat is either released or absorbed. What I truly did in this experience is to calculate this change of energy. For the first part of this experience. I determined the initial temperature of water and the final temperature of the solution of water and solid sodium hydroxide. I obtained 25.0 °C and 30 30°C respectively. From these temperatures, I found their temperature change. With the values of temperature, mass of water and specific heat of water. I calculated the heat released which gave me a value of 1109.3 J. Next, the mass of NaOH gave an amount of mole equaled to 0.025. With this valua. I was able to determine 44.4 KJ molar heat of the reaction, which is the ratio of the heat released and amount of mole calculated 2. Hess's law states that if a reaction takes place in several steps, then its standard reaction enthalpy is the sum of standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature. This experiment involved three exothermic chemical reactions. Each of this reaction had a standard enthalpy. If we add AH; to A Hš, we obtain approximatively AH- NaOH, — Naze + OH 44.4 kJ Na-2) + OH- -- + H+ (an) - C1(04) -H,00 + Nat(ar) + C1-29 +56,09 kJ NaOH) + H+(29) + Cla) — H2O + Nacea + C1 + 100.49 kJ So, we can say this experiment allowed me to calculate and compare the quantity of heat energy released in these three reactions by using Hess s law