Research question: how to convert NaOH to NaCl by two different routes , and measure the enthalpy changes for each one in order to test Hesss law ?

                         NaOH (s), 2M HCl (aq)                                                   2M NaCl (aq)

                                                                                                                           

                        H solu                                                                                      Hdilu    

                                                                                                                       

                                                                             

                                                                         

                                                                                       

                 2M NaOH (aq) + 2M HCl (aq)                                                 1MNaCl (aq)

                             50.00cm3                                                         

Uncertainties:

• to find the required amount of NaOH needed for each reaction :

Number of moles of NaOH solution = concentration x

n = 2.0 x   = 0.10 mol

M (NaOH) = 22.99 + 16.00 + 1.01 = 40.0 g mol -1

n =    which implies that    m = n x M

m(NaOH)  = 0.10 x 40.0 = 4.0 g

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• to find the mass of the water :

The amount of water to be used was given, 50.0cm3, but since

1 ml = 1g

1ml=1cm3

1cm3=1g       implying that 50.0cm3 = 50.0g

 Or  using the  function of density  :  density=        

 (It is known that the density of water = 1.00g /cm3 )

 1.00 g/cm3=             which implies that       mass of water = 1.00 x 50.0 = 50.0 g

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Table 1 : the table below shows the raw data collected when  adding 4.0 g of  sodium hydroxide NaOH  to 50.0 cm3 of  water H2O  in the thermos and measuring the temperature change for 450 seconds using labQuest when applying route (A),  :

• The bolded and underlind data presents the original temperature of the water before adding the sodium hydroxide.

    NaOH (s) + 2M H2O (l)                          2M NaOH (aq)  

Graph (1)  the graph shows the maximum theoretical temperature obtained when  of water was added to 4.0 g sodium hydroxide  NaOH including the best-fit line, (temperature change was measured every 30 seconds for 450 seconds).

From graph (1) highest temperature could be determined to be 35.0˚C

Calculations :

Table 2 : the table below shows the raw data collected when  adding 50.00 cm3 of HCl ,to the solution 4.0 g of  sodium hydroxide NaOH  and 50.0 cm3 of  water H2O  in the thermos and measuring the temperature change for 450 seconds by applying route (A),  :

• The bolded and underlined data presents the original temperature of the water and sodium hydroxide before adding the HCl

2M NaOH (aq) + 2M HCl (aq)                          1M NaCl (aq)  

Graph (2 ) : the graph shows the theoretical maximum temperature obtained when 50.00cm3 of HCl was added to the solution of 4.0 g sodium hydroxide NaOH, and 50.0 cm3  of water,  the temperature change was measured every 30 seconds(including the best-fit line).

From graph (2) highest temperature could be determined to 32.2˚C.

Calculations:

So :

  NaOH (s) + 2M  (l)                    2M NaOH (aq)     = 31  kJmol-1(±1.5 kJmol-1)

+    

  2M NaOH (aq) + 2M HCl (aq)               1M NaCl (aq)  = 54 kJmol-1 (±3.4 kJmol-1)

NaOH (s) + 2M  (l) + 2M HCl (aq)                       1M NaCl (aq)

∆H = -85 (±4.9 kJmol-1)

Table 3 : the table below shows the raw data collected when  adding 4.0 g of  sodium hydroxide NaOH  to 50.00 cm3 of  HCl in the thermos and measuring the temperature change by applying  route (B)  :       NaOH (s) + 2M (l)                          2M NaCl (aq)  

• The bolded and underlined data presents the original temperature of the HCl before adding the sodium hydroxide.

Graph (3) the graph shows temperature change when 50.00 cm3 of HCl was added to 4.0 g of sodium hydroxide (temperature change was measured every 30 seconds for 450 seconds) following route (B) including the best-fit line . 

From graph (3) highest temperature could be determined to 61.0˚C.

Calculations:

Table 4 : the table below shows the raw data collected when  adding 50.0 cm3 of water ,to the solution 4.0 g of  sodium hydroxide NaOH  and 50.00 cm3 of  hydrochloric acid HCl  in the thermos and measuring the temperature change by applying route (B),  :

• The bolded and underlined data presents the original temperature of the hydrochloric acid and sodium hydroxide solution before adding the water
• 2M NaCl (aq) + (l)                         1M NaCl (aq)

Since there is no change in temperature, ∆H = 0 kJmol-1

NaOH (s) + 2M (l)                          2M NaCl (aq)      = -79 kJmol-1 (±0.66 kJmol-1)

2M NaCl (aq) + (l)                         1M NaCl (aq)      

NaOH (s) + 2M  + (l)                        1M NaCl (aq)

  ∆H= -79 kJmol-1 (±0.66 kJmol-1)

Table 5 :

The table below shows the observations of the reactions in both route (A) and route (B) under the time when the experiment was done in order to convert NaOH to NaCl by two different routes :

Conclusion and evaluation:

Hess's Law states that the enthalpy change for a reaction is identical to the enthalpy change of that reaction by any other pathway. In other words, the sum of the enthalpy changes of the intermediate steps of a reaction are equal to the enthalpy of the overall reaction, in this experiment , Hess´s law was tested by converting NaOH to NaCl in two different routes,

Two reactions were carried out in route (A) , while the third reaction in route (B) was a result of adding the first and second reaction

In route (A) :the enthalpy change when adding adding H2O to NaOH , according to the following formula :

    NaOH (s) + 2M  (l)                          2M NaOH (aq)           =-31   (± 0.68)

The enthalpy change when adding HCl to the NaOH solution :

 2M NaOH (aq) + 2M HCl (aq)                   1M NaCl (aq)     (±1.8kJ mol-1 )

In total , the  final enthalpy change is to be = -31+ -54 = - 85 (±2.5 8kJ mol-1)

In route (B), the enthalpy change was counted only from the first step , which is adding HCl to NaOH

NaOH (s) + 2M (l)                          2M NaCl (aq)      ∆H= 80.67(±0.66 kJmol-1)

As seen above, the enthalpy change is negative, meaning that it is an exothermic reaction(energy is released  in form of heat),and to be more specified , this is a neutralization reaction and it happens when an acid and basic compounds react with each other to produce salt and water. In this experiment the hydrochloric acid (HCl is the acid) and the sodium hydroxide (NaOH)is the basic compound. The graphs (1, 2, 3)were done in order to determine the maximum  temperature allowing for heat loss since it is very hard to record the maximum temperature .  

Since Hess’s law states that the enthalpy change for any chemical reaction is independent of the route provided the starting conditions and final conditions are the same. This means that the enthalpy change of the two routes (A), (B) must equal to each other.

But by comparing the result obtained for the enthalpy change of the two routes, it can be seen that there is a huge different between the two values and the uncertainties do not cover the difference, implying that there is a systematic error.

These are astronomical percent errors. There were major mistakes with equipment and procedure. A few sources of error that could account for the discrepancies are the measuring devices themselves. A graduated cylinder was used to measure volume of water. A buret or pipet should have been used for maximum accuracy. Using a graduated cylinder leaves room for a crucial error in volume determination, which would then lead to errors in determination of mass, , and every other derivative formula. Not only could an inaccurate amount of water be poured into the cylinder, but not all of this water may be poured out. There are always extra drops of solution clinging to the walls of a cylinder. This would have negatively influenced volume readings.

Another error achieved in this experiment can be accounted for is that the thermos that was used did not have a lid meaning that a lot of heat escaped to the surroundings. Since the reaction is an exothermic reaction, energy was released , an improvement of this could be to use an isolated themos with a lid to prevent heat loss

Another error that presented itself was the use of the thermometer  sometimes both as a stirring rod as well as a measurement tool. by swirling the solutions in both route (A) (B) that created a hole in the solution, in a result temperature readings could have been inaccurate because the thermometer would not encounter that much of the solution molecules and may be take the reading of the air temperature in that hole, and therefore the final calculations are incorrect.

In addition: the NaOH  was not grind well, so there were some pieces that were relatively large, so they took longer time to dissolve which might have changed the reaction results , an improvement of this error could be to use an electrical mortar or a coffee grinder so that all NaOH is grind evently

One last point is that the thermos was washed immediately with water after each reaction , and then the next reaction was performed the thermos was not dried carefully. It is highly possible that some areas of the thermos were not washed down properly and some drops of water were left. This would have changed the outcome of the reaction as well as the enthalpy calculation at the end because it would give different temperature changes.

So the error that encountered in this experiment is systematic errors more than random errors.

The experiment was successful to the degree that the procedure was carried out correctly. Equipment problems and other unavoidable sources of error served to cause a large percent error for each part of the experiment.

Assuming that the solution has the same density and specific heat capacity as water

According to "Chemistry" by Raymond Chang (9th ed), it is -56.2 kj/mol p.243.

50.0 cm3 of

Water added

50.0 cm3 of

Water added

∆T

% error =  ( i found it to be -56kjmol-1)