Calculation on the Molar Enthalpy Change of a Neutralization Reaction

Calculation on the molar enthalpy change of a neutralization reaction
Principle H2SO4(aq)+2NaOH(aq)=Na2SO4(aq)+2H2O(l) The reaction is a exothermic reaction, when the heat released by it is absorbed by water, the temperature of the water increases. The heat produced by the reaction can be calculated if it is assumed that all the heat is absorbed by the water. Heat change of reaction=-heat change of water =-mH2O×cH2O×△TH2O As the water has gained the heat produced by the reaction, the heat change of reaction is negative when the temperature of the water increases. As the heat change observed depends on the amount of reaction, for example the number of moles of fuel burned, enthalpy change reaction are usually expressed in kJmol-1. Chemicals Name of chemicals concentration Volume Water used for diluting Materials Name of materials Measuring cylinder Measuring cylinder Beaker Stirring rod Thermometer Balance Constant temperature tank stopwatch Safety goggle

Sulfuric acid 18 mol V-1 10ml 125ml

Sodium hydroxid 12 mol V-1 15ml 105ml

Condition 10 ml 100ml 50ml

Including poise Including plastic stir bar

Number of materials used 1 2 1 1 1 1 1 1 1

Procedures 1. Dilute both solutions H2SO4 and NaOH, from the given concentration to 1M and 2M. The volumes of the diluted solutions H2SO4 and NaOH with concentration 1M are 90ml which contains 0.09 moles H2SO4 and 60 ml which contains 0.06 moles NaOH respectively. And the volumes of the diluted solutions H2SO4 and NaOH concentration 2M are 45ml which contains 0.09 moles H2SO4 and 60 ml which contains 0.12 moles NaOH respectively. 2. Package the constant temperature tank. Insert the thermometer into the top of the tank with suitable height (ensure the thermometer cannot be touched the beaker’s bottom) . 3. As the coefficient of H2SO4 and NaOH is 1:2, so the ratio of amounts of H2SO4

4. 5. 6. 7.

8.

and NaOH is 1:2. Pour 1M H2SO4 with volume of 30ml and 1M NaOH with volume of 60ml into the constant temperature tank, and then cover the top immediately. While the last step finished, record the time by using stopwatch for 180 seconds. And stir the solution by using stirring rod ceaselessly. Pour 2M H2SO4 with volume of 30ml and 2M NaOH with volume of 30 ml into the constant temperature tank, and then cover the top immediately. While the last step finished, record the time by using stopwatch. And stir the solution by suing stirring rod ceaselessly. Record the time until the temperature was almost constant (theoretically, temperature will fall to room temperature after the reaction is finished, but the insulation of the equipment is good so that the decreasing of temperature will marginally). Using the time recorded and the temperature changed as coordinate axis and draws a graph of the relationship between time and temperature.

Data collection The figure below is the time recorded and the temperature changed. When the concentration of sulfuric acid and sodium hydroxide are both 1M: Time 0 5 10 15 20 25 30 60 90 120 150 180 recorded(second) Temperature 18 21 24 26 28 29.5 29.5 29.5 29.1 29 28.8 28.7 changed(Celsius) The reaction was finished after around 25 seconds, so the temperature did not increase until 25 seconds. And when the conentration of sulfuric acid and sodium hydroxide are both 2M: Time 0 5 10 15 20 25 recorded(second) Temperature 18 23 27 31 33 35 changed(Celsius) 30 37 60 37 90 120 150 180 36

36.5 36.1 36

The reaction was finished after around 30 seconds, so the temperature did not increase any more until 30 seconds.

Temperature/calcius
40 35 30 25 20 15 10 5 0 0 50

temperature VS time

1M 2M

100

150

200

Time/second

From the figure above, it is easy to prove that it is an exothermic reaction. Calculation on molar enthalpy change: △Hsystem=0(assuming no heat loss) △Hsystem=△Hwater+△Hreaction(assuming all heat goes to the water) △Hreaction=-△Hwater For the exothermic reaction, △Hreaction is negative as heat has passed from the reation into the water. Heat transfer to water =mH2O×cH2O×△TH2O Molar heat change of reaction=-mH2O×cH2O×(no .of moles
△T(H2O) limiting reagent )

As sulfuric acid and sodium hydroxide were added without excess, either of them can be seen as the limiting reagent, I choose sodium hydroxide. Number of moles of NaOH(n)=concentration([])× Number of moles of NaOH(nNaOH)=[NaOH]× In the reaction with solutions of 1M, Molar heat change=-mH2O×cH2O×
△T(H2O) (no .of moles limiting reagent ) △T H2O NaOH ×
V NaOH 1000

volume cm 3 (V) 1000

V NaOH 1000

=-mH2O×cH2O×

=-90×10-3×4.2×103× =-69300J =-69.3kJ In the reaction with solutions of 2M,

29−18
60 1000

1×

Molar heat change=-mH2O×cH2O×(no .of moles =-mH2O×cH2O×
NaOH ×

△T(H2O) limiting reagent )

△T H2O
V NaOH 1000

=-60×10-3×4.2×103×

37−18
30 1000

2×

=-79800J =-79.8kJ Graph of molar enthalpy change of reaction with solution 1M and 2M are shown below respectively.

-60 -65 -70 -75 -80 1M 2M

The value above is the actual enthalpy change of reaction with solution of 1M. The theoretical value can be measured according to the reaction, H2SO4(aq)+2NaOH(aq)=Na2SO4(aq)+2H2O(l) △Hreaction=△Hproducts-△Hreactants By searching the heat formation data, △Hreaction can be calculated. △Hreaction=-285.8×2+(-1387)-(-909.2)-2×(-469.4) =-1211.8kJ The reason of why actual value and theoretical value are so different will be discussed in evaluation. Conclusion In conclusion, the molar enthalpy change of reaction with solution of 1M is -69.3kJ and that of 2M is -79.8kJ. And it is convinced that the as the amount of reaction increasing, the absolute value of molar enthalpy change will increases, because the more mole contains in solutions, the more bonds will be broken so that more energy will be released. Evaluation As a great different between theoretical value and actual value, there must exist several problems during operating the experiment. 1. Before the sodium hydroxide was diluted, the solid state of it was balanced first. As sodium hydroxide can easily react with water vapor and carbon dioxide in the

air in room temperature. Before two steps of weighing and diluting, sodium hydroxide was exposed to the air. So the mass of sodium hydroxide decreases, which cause the concentration of solution changed. 2. After diluting concentrated sulfuric acid, the sulfuric acid solution is exposed to air as well. As sulfuric acid is a substance with high volatility, the amount of sulfuric acid can easily removed to air so that the concentration of sulfuric acid decreases which cause an error in the result. 3. As the constant temperature tank is not transparent, the position of thermometer cannot be confirmed. So it is possible to make the thermometer to contact with the bottom of constant temperature tank. Since the temperature of solutions and tank are little different, there will be an imprecise value when reading the thermometer. 4. As the reaction is exothermic reaction, temperature increases when the reaction was operated. As the reaction occurred in a closed container, when temperature increasing, the pressure increases as well. The change of pressure definitely influenced the output of the reaction. Improvement 1. Since sodium hydroxide can react with water vapor and carbon dioxide when it is exposed to air, sodium hydroxide should be diluted as quickly as possible after it was weighed. 2. Due to the high volatility of sulfuric acid, before the solution will be waited for using a long time after diluting, sulfuric acid is supposed to cover a piece of paper in order to avoid volatize. Otherwise, use it as quickly as possible after diluting. 3. To avoid thermometer from contact with the bottom of tank, it is supposed to be prepared that measuring the depth of tank so that the thermometer can be put in a right position. References Pearson Baccalaureate Standard Level Chemistry http://faculty.ccri.edu/aahughes/GenChemII/Lab%20Experiments/Enthalpy_of_Neutralization.pdf

References: Pearson Baccalaureate Standard Level Chemistry http://faculty.ccri.edu/aahughes/GenChemII/Lab%20Experiments/Enthalpy_of_Neutralization.pdf