Short Answer
Enthalpy of the Decomposition of Ammonium Chloride
Experiment 1: Find the Calorimeter Constant
Lab Results
1. Record the following lab results in the table below.
volume of cold water in the calorimeter
initial temperature of the cold water
volume of hot water added to the calorimeter
initial temperature of the hot water
final temperature in the calorimeter 50.0 mL
21.5 C
50.0 mL
60.0 C
39.4 C Data Analysis
2. Calculate the calorimeter's constant.
Ccal = qcal/DeltaTcal
50g (4.184 j/g) (39.4-60) = -4309.52
50g (4.184 j/g) (39.4-21.5) = 3744.68
4309.52 - 3744.68 = 564.84
564.84/(39.4-21.5)= 31.6 J/C
Experiment 2: Determine the Enthalpy of Neutralization of HCl by NH3
Lab Results
1. Record the following lab results in the table below.
volume of HCl solution in the calorimeter
volume of NH3 solution added
final temperature in the calorimeter 25.0 mL
25.0 mL
33.2 C Data Analysis
2. Calculate the number of moles of ammonia and hydrochloric acid.
Ammonia 25/17=1.4706 moles
Hydrochloric Acid 25/36.5 = 0.685 moles
3. Calculate the enthalpy of the neutralization reaction between HCl and NH3.
50g (4.184 j/g x (33.2 C – 21.5 C)) = 2445.3 J
Heat absorption 21.5 j/c x 11.7 = 254 J
Add together 2699.3 J
Enthalpy of the reaction, per mole of HCl
2699.3 J/ 0.05 mol = 53,986 J per mole of HCl
Enthalpy of the reaction, per mole of NH3 (J/mol)
2699.3 J/0.050 mol = 53,986 J per mole of NH3 Short Answer
Enthalpy of the Decomposition of Ammonium Chloride
Experiment 1: Find the Calorimeter Constant
Experiment 3: Determine the Enthalpy of Dissolution of NH4Cl in Water
Lab Results
1. Record the following lab results in the table below.
volume of water added to the calorimeter
initial temperature of the water
mass of NH4Cl added to the calorimeter
final temperature in the calorimeter 25.0 ml
21.5 C
5.000 g
9.1 C Data Analysis
2. Calculate the enthalpy of the dissolution of NH4Cl in water. The molar mass of NH4Cl is
53.49 g/mol.
NH4Cl 5g/53.49=0.09347 moles
25g (4.184 j/g) (21.5C-9.1)= 1297.04 J
(21.5 j/c)x(12.4j/c)=267 J
1297.04+267 = 1564.04 J
1564.04 /0.09347 moles = 16,733 J/Mole of NH4Cl 3. Write out the reaction
as a series of steps which include the
reactions observed in experiments 2 and 3. Use the known enthalpies for the change
of state of NH3 and HCl give below.
NH4Cl(s) -> NH(g) + HCl (g)
NH3(g) NH3 (aq) = -34,640 J/mol
HCl (g) HCl (aq) = -75,140 J/mol
NH4Cl (s) NH4Cl (aq) = 53,986 Conclusions
1. Is the neutralization reaction between ammonia and hydrochloric acid exothermic or
endothermic?
endothermic
2. Is the dissolution of ammonium chloride into water exothermic or endothermic?
endothermic
3. Given the data in the table below, what is the enthalpy of dissolution of KOH? The molar
mass of KOH is 56.11 g/mol, the specific heat of solution is 4.184 J/gºC, and the
calorimeter constant is 22.45J/ºC. Short Answer
Enthalpy of the Decomposition of Ammonium Chloride
Experiment 1: Find the Calorimeter Constant
mass of water added to the calorimeter
50.000 g
initial temperature of the water
22.0 ºC
mass of KOH added to the calorimeter
1.824 g
final temperature in the calorimeter
29.9 ºC
1.824 g/56.11= 0.03251 moles
50g (4.184 j/g) (29.9-22.0)= 1652.68 J
(7.9)x(22.45)= 177 J
1652.68 + 177 = 1829.68 J
1829.68 / 0.03251 moles = 56,281 J/Mole of KOH

 

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