Enthalpy of neutralization

  1. Introduction:

Acid and base substances are identified by using a pH meter if the substance pH was lower than 7 means it is acid as it goes lesser the more acidic it gets such as: Hydrochloric acid (HCl) and Sulfuric acid (H2SO4). And if the substance pH was more than 7 means it is base such as: Sodium hydroxide (NaOH) and Potassium hydroxide (KOH) where 7 is neutral. The enthalpy of neutralization is when acid and base react and form salt plus water. Aqueous hydrogen ion (H+(aq)) from the acid and aqueous hydroxide ion (OH(aq)) from the base combine and form water energy will be released. This energy is called enthalpy of neutralization, Means the reaction is exothermic as heat was released due to the formation of bonds (BBC, n.d.).

  1.2- Aim:

In this experiment it will be studied how enthalpy changes for different neutralizations depending on the number of moles of water formed as it is different every time according to the acids and bases in each reaction.

12- Materials and Methods

    2.1-Materials:

  • Plastic beaker
  • Measuring cylinder
  • Thermometer
  • 2.0M Hydrochloric acid solution
  • 2.0M Nitric acid solution
  • 2.0M Sulfuric acid solution
  • 2.0M Sodium hydroxide solution
  • 2.0M Potassium hydroxide solution
  • 4.0M Sodium hydroxide solution

2.2- Methods:

  1. First 25cm3 of the acid was measured into the plastic beaker and its temperature was recorded.
  2. Then 25cm3 of the base was poured into the measuring cylinder and its temperature was recorded.
  3. The base was poured into the same plastic beaker the solution was stirred and its final temperature was taken.
  4. These steps were repeated for each of the following:
ExperimentAcidBase
12.0M Hydrochloric acid2.0M Sodium hydroxide
22.0M Hydrochloric acid2.0M Potassium hydroxide
32.0M Nitric acid2.0M Sodium hydroxide
42.0M Sulfuric acid2.0M Sodium hydroxide
52.0M Sulfuric acid4.0M Sodium hydroxide
62.0M Hydrochloric acid4.0M Sodium hydroxide

3- Results

  3.1- Table:

ExperimentAcidAcid’s TemperatureBaseBase’s TemperatureSolution Temperature
12.0M Hydrochloric acid20oC2.0M Sodium hydroxide20oC31oC
22.0M Hydrochloric acid20oC2.0M Potassium hydroxide20oC30oC
32.0M Nitric acid19oC2.0M Sodium hydroxide19oC29oC
42.0M Sulfuric acid20oC2.0M Sodium hydroxide18oC46oC
52.0M Sulfuric acid20oC4.0M Sodium hydroxide20oC50oC
62.0M Hydrochloric acid20oC4.0M Sodium hydroxide20oC32oC

 

 

3.2- Calculations:

 

   Experiment 1:

 

HCl + NaOH                    NaCl + H2O

 

 

            Energy released:

 

q = mcΔT

q = 50 x 4.2 x (31-20)

                                                      q = 2310 J

q = 2.31 kJ

         

Moles of water formed in the reaction:

 

Number of Moles = Volume x Concentration

25 cm3÷ 1000 = 0.025 dm3

Number of Moles = 0.025 x 2

Number of Moles = 0.05 mol

0.05 Moles of water were formed in this reaction.

 

 

          Enthalpy change of neutralization:

 

ΔH = – q ÷ Number of Moles

ΔH = – 2.31÷ 0.05

ΔH = – 46.2 kJ mol-1

 

 

 

 

 

Experiment 2:

 

HCl + KOH                   KCl + H2O

 

Energy released:

q = mcΔT

q = 50 x 4.2 x (30-20)

                                                      q = 2100 J

q = 2.1 kJ

Moles of water formed in the reaction:

 

Number of Moles = Volume x Concentration

25 cm3÷ 1000 = 0.025 dm3

Number of Moles = 0.025 x 2

Number of Moles = 0.05 mol

0.05 Moles of water were formed in this reaction.

 

         

Enthalpy change of neutralization:

 

ΔH = – q ÷ Number of Moles

ΔH = – 2.1÷ 0.05

ΔH = – 42 kJ mol-1

 

 

 

 

 

Experiment 3:

 

HNO3 + NaOH                   NaNO3 + H2O

 

Energy released:

q = mcΔT

q = 50 x 4.2 x (29-19)

                                                      q = 2100 J

q = 2.1 kJ

Moles of water formed in the reaction:

 

Number of Moles = Volume x Concentration

25 cm3÷ 1000 = 0.025 dm3

Number of Moles = 0.025 x 2

Number of Moles = 0.05 mol

0.05 Moles of water were formed in this reaction.

 

Enthalpy change of neutralization:

 

ΔH = – q ÷ Number of Moles

ΔH = – 2.1÷ 0.05

ΔH = – 42 kJ mol-1

 

 

 

 

Experiment 4:

 

H2SO4 + 2NaOH                   Na2SO4 + 2H2O

 

Energy released:

q = mcΔT

q = 50 x 4.2 x (46-19)

                                                      q = 5670 J

q = 5.67 kJ

 

Moles of water formed in the reaction:

 

Number of Moles = Volume x Concentration

25 cm3÷ 1000 = 0.025 dm3

Number of Moles = 0.025 x 4

Number of Moles = 0.10 mol

0.05 Moles of water were formed in this reaction.

 

         

Enthalpy change of neutralization:

 

ΔH = – q ÷ Number of Moles

ΔH = – 5.67÷ 0.10

ΔH = – 56.7 kJ mol-1

 

 

 

 

Experiment 5:

 

 H2SO4 + 2NaOH                   Na2SO4 + 2H2O

 

Energy released:

q = mcΔT

q = 50 x 4.2 x (50-20)

                                                      q = 6300 J

q = 6.3 kJ

 

Moles of water formed in the reaction:

 

Number of Moles = Volume x Concentration

25 cm3÷ 1000 = 0.025 dm3

Number of Moles = 0.025 x 2

Number of Moles = 0.05 mol

0.05 Moles of water were formed in this reaction.

 

 

Enthalpy change of neutralization:

 

ΔH = – q ÷ Number of Moles

ΔH = – 6.3 ÷ 0.05

ΔH = – 126 kJ mol-1

 

 

 

 

 

Experiment 6:

 

HCl + NaOH                   NaCl + H2O

 

Energy released:

q = mcΔT

q = 50 x 4.2 x (32-20)

                                                      q = 2520 J

q = 2.52 kJ

Moles of water formed in the reaction:

 

Number of Moles = Volume x Concentration

25 cm3÷ 1000 = 0.025 dm3

Number of Moles = 0.025 x 2

Number of Moles = 0.05 mol

0.05 Moles of water were formed in this reaction.

 

         

 

Enthalpy change of neutralization:

 

ΔH = – q ÷ Number of Moles

ΔH = – 2.52÷ 0.05

ΔH = – 50.4 kJ mol-1

 

 

 

 

Errors:

 

In this experiment there were some errors that might have affected the results:

 

–         Using the same measuring cylinder without washing it for each experiment which may have few drops of the solution that was in it and will result in a slight difference in the results. Rinsing the cylinder or using another one could have eliminated this error.

–         After stirring the solutions, the temperatures kept fluctuating. That being said, setting a timer for a few seconds before taking measurements.

Discussion:

 

 

         

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5-Refrences:

 

 

  • BBC. (n.d.). Water and neutral solutions – Acids and bases – National 5 Chemistry Revision. BBC Bitesize. Retrieved 27 February 2022, from https://www.bbc.co.uk/bitesize/guides/zsmgpbk/revision/3
  1. Introduction:

Acid and base substances are identified by using a pH meter if the substance pH was lower than 7 means it is acid as it goes lesser the more acidic it gets such as: Hydrochloric acid (HCl) and Sulfuric acid (H2SO4). And if the substance pH was more than 7 means it is base such as: Sodium hydroxide (NaOH) and Potassium hydroxide (KOH) where 7 is neutral. The enthalpy of neutralization is when acid and base react and form salt plus water. Aqueous hydrogen ion (H+(aq)) from the acid and aqueous hydroxide ion (OH(aq)) from the base combine and form water energy will be released. This energy is called enthalpy of neutralization, Means the reaction is exothermic as heat was released due to the formation of bonds (BBC, n.d.).

  1.2- Aim:

In this experiment it will be studied how enthalpy changes for different neutralizations depending on the number of moles of water formed as it is different every time according to the acids and bases in each reaction.

12- Materials and Methods

    2.1-Materials:

  • Plastic beaker
  • Measuring cylinder
  • Thermometer
  • 2.0M Hydrochloric acid solution
  • 2.0M Nitric acid solution
  • 2.0M Sulfuric acid solution
  • 2.0M Sodium hydroxide solution
  • 2.0M Potassium hydroxide solution
  • 4.0M Sodium hydroxide solution

2.2- Methods:

  1. First 25cm3 of the acid was measured into the plastic beaker and its temperature was recorded.
  2. Then 25cm3 of the base was poured into the measuring cylinder and its temperature was recorded.
  3. The base was poured into the same plastic beaker the solution was stirred and its final temperature was taken.
  4. These steps were repeated for each of the following:
ExperimentAcidBase
12.0M Hydrochloric acid2.0M Sodium hydroxide
22.0M Hydrochloric acid2.0M Potassium hydroxide
32.0M Nitric acid2.0M Sodium hydroxide
42.0M Sulfuric acid2.0M Sodium hydroxide
52.0M Sulfuric acid4.0M Sodium hydroxide
62.0M Hydrochloric acid4.0M Sodium hydroxide

3- Results

  3.1- Table:

ExperimentAcidAcid’s TemperatureBaseBase’s TemperatureSolution Temperature
12.0M Hydrochloric acid20oC2.0M Sodium hydroxide20oC31oC
22.0M Hydrochloric acid20oC2.0M Potassium hydroxide20oC30oC
32.0M Nitric acid19oC2.0M Sodium hydroxide19oC29oC
42.0M Sulfuric acid20oC2.0M Sodium hydroxide18oC46oC
52.0M Sulfuric acid20oC4.0M Sodium hydroxide20oC50oC
62.0M Hydrochloric acid20oC4.0M Sodium hydroxide20oC32oC

 

 

3.2- Calculations:

 

   Experiment 1:

 

HCl + NaOH                    NaCl + H2O

 

 

            Energy released:

 

q = mcΔT

q = 50 x 4.2 x (31-20)

                                                      q = 2310 J

q = 2.31 kJ

         

Moles of water formed in the reaction:

 

Number of Moles = Volume x Concentration

25 cm3÷ 1000 = 0.025 dm3

Number of Moles = 0.025 x 2

Number of Moles = 0.05 mol

0.05 Moles of water were formed in this reaction.

 

 

          Enthalpy change of neutralization:

 

ΔH = – q ÷ Number of Moles

ΔH = – 2.31÷ 0.05

ΔH = – 46.2 kJ mol-1

 

 

 

 

 

Experiment 2:

 

HCl + KOH                   KCl + H2O

 

Energy released:

q = mcΔT

q = 50 x 4.2 x (30-20)

                                                      q = 2100 J

q = 2.1 kJ

Moles of water formed in the reaction:

 

Number of Moles = Volume x Concentration

25 cm3÷ 1000 = 0.025 dm3

Number of Moles = 0.025 x 2

Number of Moles = 0.05 mol

0.05 Moles of water were formed in this reaction.

 

         

Enthalpy change of neutralization:

 

ΔH = – q ÷ Number of Moles

ΔH = – 2.1÷ 0.05

ΔH = – 42 kJ mol-1

 

 

 

 

 

Experiment 3:

 

HNO3 + NaOH                   NaNO3 + H2O

 

Energy released:

q = mcΔT

q = 50 x 4.2 x (29-19)

                                                      q = 2100 J

q = 2.1 kJ

Moles of water formed in the reaction:

 

Number of Moles = Volume x Concentration

25 cm3÷ 1000 = 0.025 dm3

Number of Moles = 0.025 x 2

Number of Moles = 0.05 mol

0.05 Moles of water were formed in this reaction.

 

Enthalpy change of neutralization:

 

ΔH = – q ÷ Number of Moles

ΔH = – 2.1÷ 0.05

ΔH = – 42 kJ mol-1

 

 

 

 

Experiment 4:

 

H2SO4 + 2NaOH                   Na2SO4 + 2H2O

 

Energy released:

q = mcΔT

q = 50 x 4.2 x (46-19)

                                                      q = 5670 J

q = 5.67 kJ

 

Moles of water formed in the reaction:

 

Number of Moles = Volume x Concentration

25 cm3÷ 1000 = 0.025 dm3

Number of Moles = 0.025 x 4

Number of Moles = 0.10 mol

0.05 Moles of water were formed in this reaction.

 

         

Enthalpy change of neutralization:

 

ΔH = – q ÷ Number of Moles

ΔH = – 5.67÷ 0.10

ΔH = – 56.7 kJ mol-1

 

 

 

 

Experiment 5:

 

 H2SO4 + 2NaOH                   Na2SO4 + 2H2O

 

Energy released:

q = mcΔT

q = 50 x 4.2 x (50-20)

                                                      q = 6300 J

q = 6.3 kJ

 

Moles of water formed in the reaction:

 

Number of Moles = Volume x Concentration

25 cm3÷ 1000 = 0.025 dm3

Number of Moles = 0.025 x 2

Number of Moles = 0.05 mol

0.05 Moles of water were formed in this reaction.

 

 

Enthalpy change of neutralization:

 

ΔH = – q ÷ Number of Moles

ΔH = – 6.3 ÷ 0.05

ΔH = – 126 kJ mol-1

 

 

 

 

 

Experiment 6:

 

HCl + NaOH                   NaCl + H2O

 

Energy released:

q = mcΔT

q = 50 x 4.2 x (32-20)

                                                      q = 2520 J

q = 2.52 kJ

Moles of water formed in the reaction:

 

Number of Moles = Volume x Concentration

25 cm3÷ 1000 = 0.025 dm3

Number of Moles = 0.025 x 2

Number of Moles = 0.05 mol

0.05 Moles of water were formed in this reaction.

 

         

 

Enthalpy change of neutralization:

 

ΔH = – q ÷ Number of Moles

ΔH = – 2.52÷ 0.05

ΔH = – 50.4 kJ mol-1

 

 

 

 

Errors:

 

In this experiment there were some errors that might have affected the results:

 

–         Using the same measuring cylinder without washing it for each experiment which may have few drops of the solution that was in it and will result in a slight difference in the results. Rinsing the cylinder or using another one could have eliminated this error.

–         After stirring the solutions, the temperatures kept fluctuating. That being said, setting a timer for a few seconds before taking measurements.

Discussion:

 

 

         

 

 

 

 

 

 

 

 

 

 

 

 

5-Refrences:

 

 

  • BBC. (n.d.). Water and neutral solutions – Acids and bases – National 5 Chemistry Revision. BBC Bitesize. Retrieved 27 February 2022, from https://www.bbc.co.uk/bitesize/guides/zsmgpbk/revision/3

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