 We use cookies to give you the best experience possible. By continuing we’ll assume you’re on board with our cookie policy

# Using Hess’s Law to Calculate Enthalpy Change Essay Sample The whole doc is available only for registered users OPEN DOC

## A limited time offer!

Get a custom sample essay written according to your requirements urgent 3h delivery guaranteed

Order Now

## Using Hess’s Law to Calculate Enthalpy Change Essay Sample

Controlled variables: – Mass of water

– Mass of magnesium sulphate anhydrous

– Mass of MgSO4.7H20

Materials:

1) Safety spectacles

2) 2 filter papers

3) Spatula

4) Digital balance

5) 2 polystyrene cups

6) Thermometer

7) Magnesium sulphate anhydrous

8) Magnesium sulphate-7- water

9) Distilled water

10) Measuring cylinder

Procedure:

Part A

1) Weigh 3.01g of MgSO4 anhydrous to the nearest 0.01g on a filter paper, using the digital balance.

2) Weigh 45.00g of water to the nearest 0.01g into a polystyrene cup using the balance.

3) Measure the initial temperature of the measured amount of water using the thermometer and record this value.

4) Add the measured amount of MgSO4 anhydrous to the water and stir. Measure and record the maximum temperature obtained using the thermometer.

Part B

1) Weigh 6.16g of MgSO4.7H20 to the nearest 0.01g on a filter paper, using the digital balance.

2) Weigh 41.85g of water to the nearest 0.01g into a polystyrene cup using the balance.

3) Measure the initial temperature of the measured amount of water using the thermometer and record this value.

4) Dissolve the MgSO4.7H20 in the water and record the temperature change associated with this process.

Results:

Mass of MgSO4 = 3.01g

Mass of Mg SO4.7 H2O = 6.16g

Mass of water in Part A = 45.00g

Mass of water in Part B = 41.85g

Part A

Part B

Initial temperature of water/ ï¿½ C

Maximum temperature of solution/ ï¿½ C

Initial temperature of water/ ï¿½ C

Minimum temperature of solution/ ï¿½ C

18.0

21.0

18.0

16.0

Discussion:

*First the enthalpy change of the reaction in part A is calculated:

(MgSO4(s) + 100H2O(l) MgSO4(aq,100H2O)

* Q = m x c x delta T

* M= 45.1 + 3.02

= 48.1g

* C = 4.18 J/g.K

* Delta T = 21 – 18

= 3.00 K

* Q = 603J

* Number of moles of water = mass / molar mass

= 45.1 / 18.0

= 2.50 moles

*Number of moles of MgSO4(s) = Mass/ Molar mass

= 3.02 / (24.31 + 32.06 + (4 x 16.00))

= 3.02 / 120

= 0.0251 moles

* Ratio of moles = 1 : 100

Thus both reactants are limiting

* Enthalpy change = Q / number of moles

= 603 / 0.0251

= – 24023.90 (rounded to 3 significant figures) = -24000 J/mol

(Exothermic reaction)

Enthalpy change for part B:

(MgSO4.7H20(s) + 93H2O(l) MgSO4(aq,100H2O))

* Q = m x c delta T

* M = 6.19 + 41.9

= 48.09 (rounded to 3 significant figures) = 48.1 g

* C = 4.18 J/g.K

* Delta T = 16 – 18

= – 2 K (place only the absolute value)

= 2.00

Q = 402.116 (rounded to 3 significant figures) = 402 J

* Number of moles of water = mass / molar mass

= 41.9 / 18.0

= 2.33 moles

*number of moles of MgSO4.7H2O = mass / molar mass

= 6.19 / (24.31 + 32.06 + (4 x 16.00) + (7x 18.02))

= 93.0 moles

* Ratio of moles = 2.33 : 0.0251

= 93.0 : 1.00

* Both reactants are limiting

* Enthalpy change = Q / number of moles

= 402/ 0.0251

= 16015.93 = 16000 J/mole (endothermic reaction)

MgSO4(s) + 7H2O MgSO4.7H2O(s)

-24000J/mole – 16000J/mole

MgSO4(aq,100H2O)

* Enthalpy change for the reaction of:

(MgSO4(s) + 7H2O MgSO4.7H2O(s) )

= -24000 – 16000

= – 40000 J/ mole

= – 40 Kj / mole

Uncertainties:

Uncertainty

Percentage uncertainty

Balance

3.02

0.005

(0.005/3.02) x 100 = 0.166%

Balance

6.19

0.005

(0.005/ 6.19) x 100 = 0.0808 %

Balance

45.1

0.005

(0.005/ 45.1) x 100 = 0.0111 %

Balance

41.9

0.005

( 0.005/ 41.9) x 100 = 0.0119 %

Thermometer

18.0

0.05

(0.05/ 18.0) x 100 = 0.278 %

Thermometer

21.0

0.05

(0.05/ 21.0) x 100 = 0.238 %

Thermometer

18.0

0.05

(0.05/ 18.0) x 100 = 0.278 %

Thermometer

16.0

0.05

(0.05/16.0) x 100 = 0.313 %

Total percentage uncertainties = 0.166 + 0.0808 + 0.0111 + 0.0119 + 0.278 + 0.238 + 0.278 + 0.313 = 1.3768 = 1.38 %

The enthalpy change = – 40 ï¿½ ((1.38/100) x 40)

= – 0 ï¿½ 0.552 Kj/ mole

Conclusion:

Percentage error: difference between calculated value and literature value divided by literature value times 100

Percentage error = ((40 – 104) / – 104) x 100

= (- 64 / 104) x 100

= 61.5 % (take absolute value)

Sources of error:

1) Heat lost to the surrounding

2) The specific heat capacity used in calculations was that of water, not the mixture itself

3) Parallax error

4) Position of thermometer

5) Stirring was not constant

Improvements:

1) Use a well insulated vessel with lid

2) Find the specific heat capacity of the mixture itself

3) Use digital thermometer connected to a computer

4) Ensure that the line of sight is perpendicular to the reading of the thermometer and measuring cylinder

5) Use magnetic stirrer for uniform stirring

6) Repeat the experiment several times, taking an average value for the heat emitted or gained by the mixture

We can write a custom essay

Order an essay 300+
Materials Daily 100,000+ Subjects
2000+ Topics Free Plagiarism
Checker All Materials
are Cataloged Well

Sorry, but copying text is forbidden on this website. If you need this or any other sample, we can send it to you via email.

Sorry, but only registered users have full access

immediately?

Become a member

Thank You A Lot! Emma Taylor

online

Hi there!
Would you like to get such a paper?
How about getting a customized one?

Can't find What you were Looking for?