Enthalpy change of neutralisation Essays

Enthalpy change of neutralisation Essays Enthalpy change of neutralisation Essay Enthalpy change of neutralisation Essay 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) à ¯Ã‚ ¿Ã‚ ½ NaCl (aq) + H2O (l) H+ + Cl- + Na+ + OH- à ¯Ã‚ ¿Ã‚ ½ Na+ + Cl- + H2O H+ + OH-à ¯Ã‚ ¿Ã‚ ½ 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. Hypothesis: If the temperature of a given substance is known, we may calculate the enthalpy of this substance. Prediction: The results of this experiment will probably be similar for sets of different acids and hydroxides. Different concentrations of the same acid will not influence the enthalpy of neutralisation. Assumptions: The density of acids is equal to the density of water and amounts to 1.00 g cm-3 Key variables: m mass of a substance in grams s specific heat capacity in J g-1 K-1 ?T the amount by which the temperature is increased in K PLANNING (B) Requirements: 1 burette (25 ml) 2 beakers 3 calibrated flasks (500 ml) 1 plastic bottle (1500 ml) phenolphthalein Procedure: We were provided with 2 mol dm-3 hydrochloric acid (HCl), 2 mol dm-3 nitric acid (HNO3), 2 mol dm-3 potassium hydroxide (KOH), 2 mol dm-3 sodium hydroxide (NaOH) and 4 mol dm-3 sodium hydroxide (NaOH). 1. We measured 30 cm3 of approximately of 2 mol dm-3 nitric acid into the beaker. 2. We took the temperature of the nitric acid and recorded it in table 1. 3. We measured 30 cm3 of approximately of 2 mol dm-3 sodium hydroxide into the beaker. 4. We took the temperature of the sodium hydroxide and recorded it in table 1. 5. Subsequently we added the NaOH to the HNO3 and stirred the mixture carefully with the thermometer. 6. While mixing we recorded the maximum temperature of the solution. 7. We repeated it 5 times with different sets of acids and hydroxides. a) hydrochloric acid and sodium hydroxide b) hydrochloric acid and potassium hydroxide c) nitric acid and potassium hydroxide d) nitric acid and sodium hydroxide e) hydrochloric acid and 4 mol dm-3 sodium hydroxide DATA COLLECTION a) HCl (aq) + NaOH (aq) à ¯Ã‚ ¿Ã‚ ½ NaCl (aq) + H2O (l) H+ + Cl- + Na+ + OH- à ¯Ã‚ ¿Ã‚ ½ Na+ + Cl- + H2O H+ + OH-à ¯Ã‚ ¿Ã‚ ½ H2O Amount of hydrochloric acid 30 cm3 Temperature of hydrochloric acid 20.0 oC Amount of sodium hydroxide 30 cm3 Temperature of sodium hydroxide 20.0 oC Amount of the mixture 60 cm3 Temperature of the mixture 31.0 oC Table 1. b) HCl (aq) + KOH (aq) à ¯Ã‚ ¿Ã‚ ½ KCl (aq) + H2O (l) Amount of hydrochloric acid 30 cm3 Temperature of hydrochloric acid 21.0 oC Amount of potassium hydroxide 30 cm3 Temperature of potassium hydroxide 20.0 oC Amount of the mixture 60 cm3 Temperature of the mixture 32.0 oC Table 2. c) HNO3 (aq) + KOH (aq) à ¯Ã‚ ¿Ã‚ ½ KNO3 (aq) + H2O (l) Amount of nitric acid 30 cm3 Temperature of nitric acid 23.0 oC Amount of potassium hydroxide 30 cm3 Temperature of potassium hydroxide 20.5 oC Amount of the mixture 60 cm3 Temperature of the mixture 33.0 oC Table 3. d) HNO3 (aq) + NaOH (aq) à ¯Ã‚ ¿Ã‚ ½ NaNO3 (aq) + H2O (l) Amount of nitric acid 30 cm3 Temperature of nitric acid 24.5 oC Amount of sodium hydroxide 30 cm3 Temperature of sodium hydroxide 21.0 oC Amount of the mixture 60 cm3 Temperature of the mixture 33.0 oC Table 4. e) HCl (aq) + NaOH (aq) à ¯Ã‚ ¿Ã‚ ½ 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) à ¯Ã‚ ¿Ã‚ ½ 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) HCl (aq) + KOH (aq) à ¯Ã‚ ¿Ã‚ ½ KCl (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) = 32.0 oC 20.5 oC = 10.5 oC heat required = ms?T = 60.0 g * 4.18 J g-1 K-1 * 10.5 oC = 2633.40 J = 2.633 kJ nHCl = c V = 2 mol dm-3 * 0.3 dm-3 = 0.06 moles nKOH = c V = 2 mol dm-3 * 0.3 dm-3 = 0.06 moles ?H = heat required * 1/n = 2.633 kJ * 1/0.06 moles = 43.88 kJ mol-1 ?H = 43.88 kJ mol-1 c) HNO3 (aq) + KOH (aq) à ¯Ã‚ ¿Ã‚ ½ KNO3 (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) = 33.0 oC 21.75 oC = 11.25 oC heat required = ms?T = 60.0 g * 4.18 J g-1 K-1 * 11.25 oC = 2821.50 J = 2.822 kJ nHCl = c V = 2 mol dm-3 * 0.3 dm-3 = 0.06 moles nKOH = c V = 2 mol dm-3 * 0.3 dm-3 = 0.06 moles ?H = heat required * 1/n = 2.822 kJ * 1/0.06 moles = 47.03 kJ mol-1 ?H = 47.03 kJ mol-1 d) HNO3 (aq) + NaOH (aq) à ¯Ã‚ ¿Ã‚ ½ NaNO3 (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) = 33.0 oC 22.75 oC = 10.25 oC heat required = ms?T = 60.0 g * 4.18 J g-1 K-1 * 10.25 oC = 2570.70 J = 2.571 kJ nHCl = c V = 2 mol dm-3 * 0.3 dm-3 = 0.06 moles nKOH = c V = 2 mol dm-3 * 0.3 dm-3 = 0.06 moles ?H = heat required * 1/n = 2.751 kJ * 1/0.06 moles = 45.85 kJ mol-1 ?H = 45.85 kJ mol-1 e) HCl (aq) + NaOH (aq) à ¯Ã‚ ¿Ã‚ ½ 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) = 33.0 oC 21.5 oC = 11.5 oC heat required = ms?T = 60.0 g * 4.18 J g-1 K-1 * 11.5 oC = 2884.20 J = 2.884 kJ nHCl = c V = 2 mol dm-3 * 0.3 dm-3 = 0.06 moles nKOH = c V = 2 mol dm-3 * 0.3 dm-3 = 0.06 moles ?H = heat required * 1/n = 2.884 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-à ¯Ã‚ ¿Ã‚ ½ 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.