ryan chem lab

8/10/2019 Ryan Chem Lab

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RRM Hrxn& Hesss Law Page 2

Procedure

For Part 1, 50.0 mL of distilled water at room temperature will need to be

measured, its initial temperature recorded, and placed into the calorimeter.

Then, 50.0 mL of heated water will need to be measured, its initial temperature

record, and placed into the calorimeter with the room temperature water. With

the motor on and the magnet spinning, immediately insert the thermometerprobe and collect the data. For Part 2, get 50.0 mL of HCl and NaOH and

record their initial temperatures. Then add them to the calorimeter, cover the

calorimeter, insert the temperature probe, and start collecting the data. Do this

procedure for Reaction 2 and Reaction 3. Then, the heat change and enthalpy

of reaction are to be calculated. The value of H is to be found two ways,

through experimental data and using Hesss Law. These values are then

compared to find the percent error of the experiment.

Data

Part 1: Determination of the Heat Capacity of the CalorimeterInitial Temperature

50.0 mL H2Oroom temperature: 22.4 C

50.0 mL H2Oheated: 68.8 C

Mixing Data

Time (seconds) Temperature (C)

20 44.1

40 43.9

60 43.7

80 43.5

100 43.4

120 43.4

140 43.4

160 43.2

180 43.0

Calculated Values

Tmix= 44.04 C Tavg= 45.6 C qcal= +652 J Ccal= 30.1 J/C

Part 2: Determination of Heats of Reaction

Reaction 1

Initial Temperature

50.0 mL 2.0 M HCl 22.9 C

50.0 mL 2.0 M NaOH 22.7 CMixing Data


20 35.8

40 35.8

60 35.8

80 35.8


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100 35.7

120 35.7

140 35.6

160 35.6

180 35.6

Calculated valuesTmix= 35.87 C qrxn= -6020 J H = -60 kJ/mol

Reaction 2

Initial Temperature

50.0 mL 2.0 M NH4Cl 23.0 C

50.0 mL 2.0 M NaOH 22.8 C

Mixing Data


20 23.8

40 23.8

60 23.880 23.8

100 23.8

120 23.8

140 23.8

160 23.8

180 23.8

Calculated Values

Tmix= 23.77 C qrxn= -401 J H = -4.0 kJ/mol

Reaction 3

Initial Temperature50.0 mL 2.0 M NH3 22.8 C

50.0 mL 2.0 M HCl 22.4 C

Mixing Data


20 44.1

40 43.9

60 43.7

80 43.5

100 43.4

120 43.4140 43.3

160 43.2

180 43.0

Calculated Values

Tmix= 27.42 C qrxn= -2160 J H = -22 kJ/mol


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Calculations

Part 1

1.

Tmix= 44.04 C (from graph)

2.

3.

( )

4.

Part 2

Reaction 1

1. Tmix= 35.87 C (from graph)

2.

[

]

3.

Reaction 2

*See calculation answers in the Data section

Reaction 3


Part 3

1.

The net ionic equations of the three chemical equations used throughout

the experiment

Equation 1:NaOH(aq)+ HCl(aq)NaCl(aq)+ H2O(l)Na+(aq)+ OH-(aq)+ H+(aq)+ Cl-(aq)Na+(aq)+ Cl-(aq)+ H2O(l)*OH-(aq)+ H+(aq)H2O(l)*Equation 2:NH4Cl(aq)+ NaOH(aq)NH3(aq)+NaCl(aq)+ H2O(l)NH4+(aq)+ Cl-(aq)+ Na+(aq)+ OH-(aq) NH3(aq)+ Na+(aq)+ Cl-(aq) + H2O(l)


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*NH4+(aq)+ OH-(aq)NH3(aq)+ H2O(l)*Equation 3:NH3(aq)+ HCl(aq)NH4Cl(aq)NH3(aq)+ H+(aq) + Cl-(aq)NH4+(aq)+ Cl-(aq)*NH3(aq)+ H+(aq)

NH4+(aq)*

***Equation 1Equation 2(reversed) = Equation 32.

Equation 1 H = -60. kJ/mol

Equation 2 H = -4.0 kJ/mol

OH-(aq)+ H+(aq)H2O(l) H = -60. kJ/molNH3(aq)+ H2O(l)NH4+(aq)+ OH-(aq) H = +4.0 kJ/molNH3(aq)+ H+(aq)NH4+(aq) H = -56 kJ/mol

Error Analysis

| | || There were many errors that could have affected our results in thisexperiment but three main errors were all of the initial temperatures were

different, the group had to assume the densities of the solutions, and the group

had to assume that the molarity of all solutions was 2.0 M.

The first error was that the initial temperature measurements for all

solutions were not the same. For example, in the first reaction of Part 2, the initial

temperature of hydrochloric acid was 22.9 C and the initial temperature of

sodium hydroxide was 22.7 C. In order to find the initial temperature, the

average of these two numbers had to be taken, which could have caused the

calculations to be slightly higher or slightly lower than what they should have

been. However, this error would have a negligible effect on the data since

there is not a large difference between the numbers.The second error that occurred in the experiment was that the group had

to assume the densities of the solutions. In the calculation selection of the lab

handout, it stated to assume the density of the solutions is 1.03 g/mL. Instead of

assuming this value, the group could have calculated it to have a more

accurate answer. Like the error before, it would have a small effect on the data

that was produced.

The third error that occurred in the experiment was that the group

had to assume that the molarity of all of the solutions were 2.0 M.

All of these solutions were student prepared and could have been

made improperly since they were the first solutions the class made.Mixing solutions that are not close to 2.0 M would have reduced the

overall molarity of the mixed solutions. If the solutions molarity was

less than 2.0 M, then it would increase the final answer, but if the

solutions molarity was greater than 2.0 M, then the final answer

would have decreased. This error would have been the biggest

concern during the lab since it had the largest effect on the data.


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Conclusion

The purpose of this experiment was not met because the group wasunable to verify Hesss Law. From the calculations, H3 was found to be -22

kJ/mol, but when Hesss Law was used, H3 was found to be -56 kJ/mol. The

percent error calculated from these values ended up being -61%. Since the

calculated answers were not nearly as close as they should have been, thegroup was unable to verify Hesss Law.

Questions

1.

The direct method worked better to find H3 than the indirect method

because in order to use Hesss Law, the data had to be nearly error free.

If the group does have an error in the calculations, than the Hesss Law

answer will not match the answer the group calculated. The Hesss Law

answer had a -61% error compared to the calculated answer. Therefore,

the direct method worked better than the indirect method.

2.

Hesss Law states that if a reaction is carried out in a number of steps, H

for the overall reaction is equal to the sum of the

Hs from eachindividual step.3.

H means the heat or enthalpy change for a chemical reaction. This

energy change is equal to the amount of heat transferred, at constant

pressure, in the reaction. This change represents the difference in

enthalpy of the products and the reactants and is independent of the

steps in going from reactants to products.

4.

The true initial temperature is found by using a modified linear fit because

when excluding any points that are skewed, it is giving the results a more

accurate reading. The first couple points that were skewed are due to

incomplete mixing and lack of equilibrium with the thermometer.

5.

All of the solutions did not have the same initial temperature becausethey were not contained in the same area of the room. These areas

could have a slightly different temperature causing the different initial

temperatures. Also, transfer of heat from holding the graduated cylinder

could have increased the initial temperature, and depending on how

long the graduated cylinder was held, could have determined how high

the initial temperature rose.


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Calculations:

Part 15. Tmix= 44.5 C (from graph)

6. 7. ( )

8.

Part 2

Reaction 14. Tmix= 36.82 C (from graph)

5.

[

]

6. Reaction 2

*See calculation answers in the Data sectionReaction 3


Part 33. The net ionic equations of the three chemical equations used throughout the experiment

Equation 1:

NaOH(aq)+ HCl(aq)NaCl(aq)+ H2O(l)Na

+(aq)+ OH

-(aq)+ H

+(aq)+ Cl

-(aq)Na+(aq)+ Cl-(aq)+ H2O(l)


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*OH-(aq)+ H

+(aq)H2O(l)*

Equation 2:

NH4Cl(aq)+ NaOH(aq)NH3(aq)+NaCl(aq)+ H2O(l)NH4

+(aq)+ Cl

-(aq)+ Na

+(aq)+ OH

-(aq) NH3(aq)+ Na+(aq)+ Cl-(aq) + H2O(l)

*NH4+

(aq)+ OH-(aq)

NH3(aq)+ H2O(l)*

Equation 3:NH3(aq)+ HCl(aq)NH4Cl(aq)NH3(aq)+ H

+(aq) + Cl

-(aq)NH4+(aq)+ Cl-(aq)

*NH3(aq)+ H+

(aq)NH4+(aq)****Equation 1Equation 2(reversed) = Equation 3

4. Equation 1 H = -60. kJ/molEquation 2 H = -4.0 kJ/mol

OH-(aq)+ H

+(aq)H2O(l) H = -60. kJ/mol

NH3(aq)+ H2O(l)NH4+(aq)+ OH-(aq)H = +4.0 kJ/molNH3(aq)+ H

+(aq)NH4+(aq) H = -56 kJ/mol

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