Determine the Enthalpy of Neutralisation for the following there Acids, H2SO4, HNO3 and H2SO4

Apparatus

• Polystyrene cup
• Glass Rod
• Thermometer
• Measuring Cylinder
• 1 moldm-3 of HNO3
• 0.5 moldm-3 of H2SO4
• 1 moldm-3 of HCl
• 1 moldm-3 of NaOH
• Stop clock
• Pen Paper, graph paper
• Ruler
• Calculator

Method

The heat released during a neutralisation (when 1 mole of water is formed) can be found buy measuring the temperature rise produced in a calorimeter. In this experiment I will be using a simple calorimeter made of a polystyrene with a lid. ‘Expanded polystyrene is a good thermal insulator and the temperature rise that occurs in the cup can be measured before the loss of heat to the surroundings becomes serious’*1  . I will carry out the neutralisation reaction in the cup and taking regular temperature readings. Use this to plot a Temp vs time graph. By extrapolating the graph find out the temperature rise use this and the E= mc∆t to calculate the energy released to the water and hence deduse the enthalpy of neutralisation.

 The diagram below shoes how I will set it up:

 

Method:

1. Using a measuring cylinder, measure 25.0cm of Acid
2. Place 25.0cm3 of Acid in the polystyrene cup, measure the temperature with a thermometer each minute for 4 minutes, using a stop clock. ( I am measuring the temperature of the acid a couple of times before I add the NaOH so that I get an accurate initial temperature as this smoothes out any fluctuations in the temperature)
3. Using a measuring cylinder measure 25.0cm of NaOH
4. After the fourth minute add  NaOH carefully and stir fully with a glass rod. Carefully take the temperature reading of the solution every minute until the 11th minute
5. Use this to plot at temperature-vs- time graph.
6. Repeat the experiment for the next acids.  

Safety

• All the acids especially H2SO4 are toxic and skin irritants and are the main hazard in the experiment. First of all I will be careful in pouring the acid so that It does not go onto my skin when in contact with acids the skin will irritate
• If the acid gets into contact with the eyes it will be difficult to remove the acid, which would then cause serious damage to the eyes.  So for this reason I will wear safety goggles at all times during the experiment, or if any explosions occur I will not damage my eyes.

 

Using the results

• Plot the temperature(°C) -vs- time(min) graph.
• Use the diagram to draw a line of best fit for the temperature of the solution before addition of alkali and draw the line of Best Fit after addition of alkali as shown in the diagram below

• Use this to find the initial temperature and the maxixmum temperature, hence the temperature achange

Obtaining Results:

 Results Tables

HCl

Conc: 1 moldm-3

Amount of acid added: 25cm3

Final temperature: 27.4°C

∆T :   5.9°C

HNO3

Conc: 1 moldm-3

Amount of acid added: 25.5

Final temperature:25.9°C

∆T :   5.9°C

                                

                                                     

H2SO4

Conc: 0.5 moldm-3

Amount of acid added: 25

Final temperature: 26.7°C

∆T :  6.2°C

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Analysis  

Calculations:

HCl

 Heat Transfer (E)  = Mass x specific heat capacity x temperature change

 Heat = m x C x ∆T

 Where:  m= mass of acid + mass of alkali

               C= specific Heat capacity 4.2Jg k.

               ∆T= temperature change (maximum temperature- initial temp)

 

We assume that the specific heat capacity of water is the same as the acid and NaOH

 Also that the 1cm3 of Acid or NaOH is = 1g

  Mass = (25 + 25)

            = 50., mass

  From the graph

  ∆T= 5.9°C

 E = 50g x 4.2 Jg k. x 5.9°C

     =  -1239.00J

  Moles of HCL = volume (dm3)   x    concentration (moldm-3)

                           = 25/1000 x 1.0

                           = 0.025 moles

  Standard Molar Enthalpy of neutralisation=  -1.239J/0.025

                                                                      =  -49.560 kJ mol-1                

HNO3

Heat Transfer (E)  = Mass x specific heat capacity x temperature change

  From the graph

  ∆T= 5.9°C

E = 50 g 4.2 Jg k. x 5.9°C

   = -1239.00J

                             

 Moles of HNO3= volume (dm3)   x    concentration (moldm-3)

                           = 25/1000 x 1.0

                           = 0.025 moles

                         

  Standard Molar Enthalpy of neutralisation=  -1.239J/0.025

                                                                      =  -49.560 kJ mol-1        

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H2SO4

Heat Transfer (E)  = Mass x specific heat capacity x temperature change

  From the graph

  ∆T= 6.2°C

E = 50 g 4.2 Jg k. x 6.2°C

  = 1302.00 J

Moles of HNO3= volume (dm3)   x    concentration (moldm-3)

                           = 25/1000 x 0.5

                           = 0.0125 moles

Unlike the other acids, the molar ratio of H2SO4 with NaOh is 1:2, shown by the following neutralisation reaction

H2SO4  +  2NaOH → Na2SO4  +  2H2O

Therefore the no. of mols. of H2SO4 = 0.0125 x  2

                                                            = 0.025mols

Standard Molar Enthalpy of neutralisation=  -1.302J/0.025

                                                                      =  -52.08 kJ mol-1        

 

As expected the Standard Molar Enthalpy of neutralisation were fairly similar HCl  HNO3  H2SO4  their    Enthalpy of neutralisation were           -49.560,  -49.560 and -52.08 kJ mol-1.  this supports my theory that Enthalpy of neutralisation of strong acids are the same, and for dibasic acid H2SO4  release around twice the amount of energy in neutralisation compared to HCl and HNO3.

However only to an extent does this support the theory as you can see with the results the Enthalpy of neutralisation H2SO4  was different to the   Enthalpy of neutralisation  for the other acids. However I believe this is more to do with inaccuracies in my experiment, and the different behaviour of diff acids.

Evaluation:

Looking at my results I can see there are some slight inaccuracies and inconsistencies. Firstly the inaccuracy highlighted above and also the values for the Enthalpy of neutralisation are slightly lower compared to values of Enthalpy of neutralisation in the textbook. I believe these are due to several errors in my experiment

1. Firstly the low values of Enthalpy of neutralisation which are result of heat being lost the surroundings before the temperature rise has been recorded. This is probably due to the fact that I didn’t use a lid on the polystyrene cup. If I was to repeat the experiment I will make sure I do this to keep in as much heat energy in the cup.
2.  Secondly my stirring for the diff acid experiment was not consistence. I may of stirred it differently for all the others, this would mean that more or less heat would of escaped, then required, which affect my thermometer readings. There are
3. Thirdly In my attempt to reduce the time needed to complete my experiment I didn’t rinse the measuring cylinder which I was using to measure out the volumes of Acid. Of course this must have led to some contamination of different acids with the acid I was measuring. Next time I will ensure that I wash all apparatus and rinse in appropriate solutions each time I repeat the experiment for different acid.

Lastly the thermometer I was using was inaccurate by o.5C. The faulty thermometer would have no effect on the calculated

enthalpy of neutralization because the thermometer read 0.50 degrees low

consistently so it would have had no effect of the ∆T, however next time I will make sure I use a thermometer which is correct to avoid any confusion.

Bibliography and References

*1 : Taken from p9 194 of Advanced Level Chemistry for AQA: A2 Student    

*2:  Taken from Ramsden A level Chemistry Book

• A-Level Chemistry (A-Level Chemistry S.)
  by E.N. Ramsden  ISBN: 0748752994#

• Chemistry AS Level and A Level (Cambridge Advanced Sciences) by Brian Ratcliff (Series Editor), Helen Eccles, John Raffan, ISBN: 0521544718

• A-level Study Guide Chemistry ('A'Level Study Guides) by Bob McDuell ISBN: 1857583361