Additivity of Heats of Reaction: Hess\'s Law 7. the initial readings in the dete

Question

Additivity of Heats of Reaction: Hess's Law 7. the initial readings in the determine the initial the final temperature, click the Statistics The maximum temperature is listed in the statistics box on the graph. Record ti and in your data table. 8. Rinse and dry the Probe, styrofoam cup, and stirring rod. Dispose of the solution as directed by your instructor. Reaction 2 9. Repeat Steps 3-8 using 100.0 mL of 0.50 M hydrochloric acid, HCl,instead of water. CAUTION: Handle the HCl solution and NaoH solid with care. Reaction 3 10. Repeat Steps 3-8, initially measuring out 50.0 mL of 1.0 M HC ead of water) into the Styrofoam calorimeter. In Step 5, instead ofsolid H measure 50.0 mL of 1.0 M NaOH solution into a graduated cylinder. After t, has been determined for the 1.0 M HCl, add the 1.0 M NaOH solution to the Styrofoam cup. CAUTION: Handle the HCl and NaoH solutions with care. PROCESSING DATA 1. Determine the mass of 100 mL of solution for each reaction (assume the density ofeach solution is 1.00 g/mL). 2. Determine the temperature change, Ar, for each reaction. 3. Calculate the heat released by each reaction, q, by using the formula: 4.18 J/g°C) 4. Find AH (AH -q). 5. Calculate moles of NaOH used in each reaction. In Reactions 1 and 2, this can be found from the mass of the NaoH. In Reaction 3, it can be found using the molarity, M, of the NaoH and its volume, in L. 6. Use the results of the Step 4 and Step 5 calculations to determine AHImol NaOH in each of the three reactions, 7. To verity the results of the experiment, combine the heat of reaction (AHImo) for Reaction l and Reaction 3. This sum should be similar to the heat of reaction (AHmol) for Reaction 2. Using the value in Reaction 2 as the accepted value and the sum of Reactions 1 and 3 as the experimental value, find the percent error for the experiment. 18-3

Explanation / Answer

Processing Data

1) Mass of 100ml of solution for each reaction (density of each solution is 1.00 g/ml)

Density and volume of each solution is same, so their mass is also same. We can calculate the mass by following formula.

M = V x d = 100 ml x 1.00 g/ml = 100 g

2) Temperature change for each reaction –

t for reaction 1 –

t = final temperature – initial temperature

    = 24.40C – 19.60 C = 4.80 C

t for reaction 2 –

t = final temperature – initial temperature

    = 33.50C – 21.90 C = 11.60 C

3) Heat released by each reaction –

Rea 1 –

q = Cp•m•Dt (Cp = 4.18 J/g°C)

= 4.18*102.201*4.8

=2046.94 J = 2.05 KJ

Rea 2 –

q = Cp•m•Dt (Cp = 4.18 J/g°C)

= 4.18*102.021*11.6 = 4946.79 J = 4.95 KJ

4) H for each reaction

Rea 1 –

H = -q

=-2.05 kJ

. Rea 2 –

H = -q

=-4.95 kJ

5) Moles of NaOH -

For rec1. 2.201*(Imol/40g)= 0.055mol   rec2. 7.04*(Imol/40g =0.176 mole rea3. M = mol/L = 0.05 mol

6) H/mol          -2.05 KJ/0.055mol =- 37.27 KJ/mol          rec2- 4.95 KJ/0.176 =-28.12KJ/mol

Rec3, -61.86kJ/mol

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