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Enthalpy change of neutralisation.

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Topic: Enthalpy change of neutralisation. I familiarised myself with the Material Safety Data Sheets of toxic substances. PLANNING (A) Enthalpy (H)1 - The sum of the internal energy of the system plus the product of the pressure of the gas in the system and its volume: Esys is the amount of internal energy, while P and V are respectively pressure and volume of the system. To measure the enthalpy we have to first figure out the mass of a substance under a constant pressure and determine the internal energy of the system. The enthalpy change (H)2 is the amount of heat released or absorbed when a chemical reaction occurs at constant pressure. The standard enthalpy change of neutralization3 is the change in enthalpy that occurs when an acid and base undergo a neutralization reaction to form one mole of water under standard conditions (298k and 1atm), i.e. react to produce water and a salt. It is a special case of the standard enthalpy change of reaction. HCl (aq) + NaOH (aq) � NaCl (aq) + H2O (l) H+ + Cl- + Na+ + OH- � Na+ + Cl- + H2O H+ + OH-� H2O Heat energy = ms?T. The amount of reat required will depend on how much of the substance there is to heat, what is it made of and the amount by which the temperature is increased. ...read more.


e) HCl (aq) + NaOH (aq) � NaCl (aq) + H2O (l) Amount of hydrochloric acid 30 cm3 Temperature of hydrochloric acid 20.5 oC Amount of 4 mol dm-3 sodium hydroxide 30 cm3 Temperature of 4 mol dm-3 sodium hydroxide 22.5 oC Amount of the mixture 60 cm3 Temperature of the mixture 33.0 oC Table 5. DATA PROCESSING AND PRESENTATION Heat required = ms?T m =d V n = c V ?T = Tmix - (T1 + T2) ?H = heat required * 1/n s = 4.18 J g-1 K-1 The amount of heat required to heat the water can be calculated as follows (we assume that the heat energy required to change the temperature of the other substances present may be ignored): a) HCl (aq) + NaOH (aq) � NaCl (aq) + H2O (l) V = 60 cm3 d = 1.00 g cm-3 m = d V = 60 cm3 * 1.00 g cm-3 = 60 g ?T = Tmix - 1/2(T1 + T2) = 31.0 oC - 20.0 oC = 11.0 oC heat required = ms?T = 60.0 g * 4.18 J g-1 K-1 * 11.0 oC = 2758 J = 2.758 kJ nHCl = c V = 2 mol dm-3 * 0.3 dm-3 = 0.06 moles nNaOH = c V = 2 mol dm-3 * 0.3 dm-3 = 0.06 moles ?H = heat required * 1/n = 2.758 kJ * 1/0.06 moles = 45.97 kJ mol-1 ?H = - 45.97 kJ mol-1 b) ...read more.


kJ * 1/0.06 moles = 48.07 kJ mol-1 ?H = - 48.07 kJ mol-1 CONCLUSION AND EVALUATION As we can see from the results above, the prediction made at the very beginning of this lab was correct. Neither type of acid or base nor the concentration of acid does not have influence on the enthalpy of neutralisation. Hence we may assume that the enthalpy of neutralisation is equal to the enthalpy change for H+ + OH-� H2O. The enthalpy change for this reaction, however, is -57.9 kJ mol-1. The differences between my results and the theoretical value may come from the fact that the measurements were not very accurate. The temperatures of the acids, bases and mixtures might have been influenced by cool beakers. Therefore the temperatures were a bit lower than they should have been. If the ?T was higher by 3oC, the enthalpy of neutralisation would be almost the same as in the sources. I do not know how to improve the experiment so that data gathered will be similar to theoretical values. I reckon in classroom conditions such mistake is not a serious one. SOURCES: 1. Green J, Damji S. 2001. Chemistry. Second edition. IBID Press, Victioria, Australia. 2. http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_neutralisation 1 The definition comes from http://www.chem.tamu.edu/class/majors/tutorialnotefiles/enthalpy.htm 2 The definition comes from http://www.ausetute.com.au/enthchan.html 3 The definition comes from http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_neutralisation ENTHALPY CHANGE OF NEUTRALISATION 1 ...read more.

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