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To investigate the effect of concentration on the temperature rise, heat evolved and heat of neutralization for the reaction between HCl and NaOH.

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Introduction

Aim: - To investigate the effect of concentration on the temperature rise, heat evolved and heat of neutralization for the reaction between HCl and NaOH. Neutralization is the special name given for the reaction between an acid and an alkali leading to the formation of water molecules and a salt. The reactions between basic oxides, or hydroxides, and acids are very important and are called neutralizations. Since the metallic ions and anions from the acid do not change, the essential reaction of neutralization is always the formation of non-ionized molecules of water from the hydroxide and hydrogen ions. H+ (aq) + OH- (aq) H2o (l) The following are examples of neutralizations. The metallic ions and the negative ions from the acids remain to produce the salt. 1/2 H2+So4- (aq) + 1/2 Na+OH- (aq) � 1/2 Na+So4- (aq) + H2o (l) Sulphuric acid Sodium hydroxide Sodium sulphate water H+No3- (aq) + Na+OH- (aq) � Na+No3- (aq) + H2o (l) Nitric acid Sodium hydroxide Sodium nitrate water H+cl- (aq) + Na+OH- (aq) � Na+Cl- (aq) + H2o (l) Hydrochloric acid Sodium hydroxide Sodium chloride water Since all these reactions reduce to H+ + OH- � H2o, the energy change accompanying both should be the same. Because, in all the three reactions only 1 mole of water is produced, and the metallic ions and the negative ions in the acid are rather spectator ions and do not participate in the reaction so much as to give an energy change. If aqueous acidic solutions are made up containing one mole of 1/2 H2So4, HNo3, and HCl in 25 cm3, and alkaline solutions containing one mole of NaOH in 25 cm3 (strong acids and alkalis), any neutralization between these solutions produces 1 mole of water and liberates the same amount of heat energy and this heat is called Heat of neutralization which is usually - 57.3 kJ. If a bond is broken, energy is needed and the reaction is endothermic. ...read more.

Middle

I noted down both the experiments' readings and found the average of those both readings. I repeated this procedure for all the three concentrations. I was so conscious about the accuracy of my test, so that I took the following precautions. Major Precautions * Stir the solution constantly without hitting the walls of the polystyrene cup. Stirring helps to spread the heat evenly to all the parts of the solution. But when the thermometer, which acts as a stirrer, hits the walls of the cup, small amount of heat is produced. This extra heat may affect my experiment. So I stirred the solution constantly without striking the walls of the container. * Measure the temperature quickly. Since the room temperature is vastly lower than the heat evolved during the reaction, there is a great chance of heat escaping to the atmosphere. So to get precise results I measured the temperature as quickly as I finished pouring the particular volumes of alkali. * Thermometer readings have to be read properly. Since I was stirring with the thermometer which actually used to measure the temperature, I was extra careful about the measurement of temperature from the thermometer. I made the thermometer stand erect and saw the temperature mark in eye level. These are the results I obtained for 1 M concentration of acid and alkali. Volume (cm3) Exp 1 - Final Temperature (�C) Exp 2 - Final Temperature (�C) Average Temperature (�C) 0.00 26.00 26.00 26.00 5.00 27.80 28.20 28.00 10.00 29.00 30.00 29.50 15.00 30.30 30.50 30.40 20.00 31.00 32.00 31.50 25.00 31.80 32.20 32.00 30.00 31.60 32.20 31.90 35.00 31.50 31.50 31.50 40.00 31.00 31.00 31.00 45.00 30.70 30.50 30.60 50.00 30.00 30.20 30.10 These are the results I obtained for 2 M concentration of acid and alkali. Volume (cm3) Exp 1 - Final Temperature (�C) Exp 2 - Final Temperature (�C) Average Temperature (�C) ...read more.

Conclusion

* Concentration is directly proportional to the heat evolved, at constant volume. * Concentration does not affect the heat of neutralization. I think that the success of my experiment solely depended on the accuracy of the experiment and I managed well enough to perform my experiment with great accuracy. I did my experiment twice for each concentration and got the average readings as the final readings and this method of doing helped me in disguising the error I might have done in one experiment. % Accuracy % error = ((Predicted value - experimental value) / predicted value) X 100 % error for temperature change in 1 M concentration = ((6.79 - 6.5) / 6.79) X 100 = (0.29 / 6.79) X 100 = 4.21 % % error for temperature change in 2 M concentration = ((13.57 - 12.5) / 13.57) X 100 = (1.07 / 13.57) X 100 = 7.89 % % error for temperature change in 3 M concentration = ((20.36 - 19.54) / 20.36) X 100 = (0.82 / 20.36) X 100 = 4.01 % % error for heat evolved in 1 M concentration = ((1425 - 1324.05) / 1425) X 100 = (100.95 / 1425) X 100 = 7.08 % % error for heat evolved in 2 M concentration = ((2850 - 2559.38) / 2850) X 100 = (290.63 / 2850) X 100 = 10.2 % % error for heat evolved in 3 M concentration = ((4275 - 3980.3) / 3980.3) X 100 = (294.7 / 3980.3) X 100 = 6.89 % % error for heat of neutralization in 1 M concentration= ((57.3 - 52.96) / 57.3) X 100 = (4.34 / 57.3) X 100 = 7.57 % % error for heat of neutralization in 2 M concentration= ((57.3 - 51.19) / 57.3) X 100 = (6.11 / 57.3) X 100 = 10.67 % % error for heat of neutralization in 3 M concentration= ((57.3 - 53.07) / 57.3) X 100 = (4.23 / 57.3) X 100 = 7.38 % ...read more.

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