Enthalpy change of neutralization = energy / no of mole of water formed
Assume: 1.) 1 cm3 = 1 g for the calculations of the six reactions below.
2.) Specific heat capacity of acid & alkali = 4200 J Kg-1K-1
3.) Specific heat capacity of glass = 840 J Kg-1K-1
4.) Specific heat capacity of plastic beaker = 0 J Kg-1K-1
Reaction 1:
Mass of the solution = (25 + 25) /1000 = 0.05 Kg
Heat given out = (0.05 * 4200 + (3/29 * 0.02073 + 3/26 * 0.02234) * 840) * 12.0
= 2567.405812 J
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
No of mole of NaOH = 2 * 0.025 = 0.05 mol
According to the equation, 1 mole of HCl reacts with 1 mole of NaOH to form 1 mole of H2O. So, No of mole of H2O = 0.05 mol
Heat given out per mole of H2O formed = 2567.405812 / 0.05 = 51.3 KJ mol-1
The enthalpy change of neutralization is –51.3KJ mol-1
Reaction 2:
Mass of the solution = (25 + 25) /1000 = 0.05 Kg
Heat given out = (0.05 * 4200 + (3/29 * 0.02073 + 3/26 * 0.02234) * 840) * 12.8
= 2738.772826 J
HCl (aq) + KOH (aq) → KCl (aq) + H2O (l)
No of mole of NaOH = 2 * 0.025 = 0.05 mol
According to the equation, 1 mole of HCl reacts with 1 mole of KOH to form 1 mole of H2O. So, No of mole of H2O = 0.05 mol
Heat given out per mole of H2O formed = 2738.772826 / 0.05 = 54.8 KJ mol-1
The enthalpy change of neutralization is –54.8 KJ mol-1
Reaction 3:
Mass of the solution = (25 + 25) /1000 = 0.05 Kg
Heat given out = (0.05 * 4200 + (3/29 * 0.02073 + 3/26 * 0.02234) * 840) * 10.2 = 2245.741176 J
HNO3 (aq) + NaOH (aq) → NaNO3 (aq) + H2O (l)
No of mole of NaOH = 2 * 0.025 = 0.05 mol
According to the equation, 1 mole of HNO3 reacts with 1 mole of NaOH to form 1 mole of H2O. So, No of mole of H2O = 0.05 mol
Heat given out per mole of H2O formed = 2245.741176 / 0.05 = 44.9 KJ mol-1
The enthalpy change of neutralization is –44.9 KJ mol-1
Reaction 4:
Mass of the solution = (25 + 25) /1000 = 0.05 Kg
Heat given out = (0.05 * 4200 + (3/29 * 0.02073 + 3/26 * 0.02234) * 840) * 13.0
= 2781.679552 J
H2SO4 (aq) + 2NaOH (aq) → Na2SO4 (aq) + 2H2O (l)
No of mole of NaOH = 2 * 0.025 = 0.05 mol
According to the equation, 1 mole of H2SO4 reacts with 2 mole of NaOH to form 2 mole of H2O. So, No of mole of H2O = 0.05 mol
Heat given out per mole of H2O formed = 2781.679552 / 0.05 = 55.6 KJ mol-1
The enthalpy change of neutralization is –55.6 KJ mol-1
Reaction 5:
Mass of the solution = (25 + 25) /1000 = 0.05 Kg
Heat given out = (0.05 * 4200 + (3/29 * 0.02073 + 3/26 * 0.02234) * 840) * 30.0 = 6419.260504 J
H2SO4 (aq) + 2NaOH (aq) → Na2SO4 (aq) + 2H2O (l)
No of mole of NaOH = 4 * 0.025 = 0.1 mol
According to the equation, 1 mole of H2SO4 reacts with 2 mole of NaOH to form 2 mole of H2O. So, No of mole of H2O = 0.1 mol
Heat given out per mole of H2O formed = 6419.260504/ 0.1 = 64.2 KJ mol-1
The enthalpy change of neutralization is –64.2 KJ mol-1
Reaction 6:
Mass of the solution = (25 + 25) /1000 = 0.05 Kg
Heat given out = (0.05 * 4200 + (3/29 * 0.02073 + 3/26 * 0.02234) * 840) * 13.5
= 2888.667227J
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
No of mole of NaOH = 4 * 0.025 = 0.1 mol
According to the equation, 1 mole of H2SO4 reacts with 2 mole of NaOH to form 2 mole of H2O. So, No of mole of H2O = 0.1 mol
Heat given out per mole of H2O formed = 2888.667227 / 0.1 = 28.9 KJ mol-1
The enthalpy change of neutralization is –28.9 KJ mol-1
Discussion
In the experiment and calculation, there are several assumptions. Firstly, we assume that there is no heat loss to the surroundings and the experiment was implemented under room condition. In addition, heat capacity of the apparatus was negligible and the specific heat capacity of the product solution is 4200 J Kg-1 K-1. Moreover, we assume 1 g of the solution equals to 1 ml of the solution.
Several sources of error are in the experiment and we can improve the experiment from such errors. At first, as the temperature of solution recorded by thermometer was increased quickly, it is difficult for us to use human eyes to get the temperature. As a result, the maximum temperature we got may be different from the actual one. So the enthalpy changes of the formation of the substances are inaccurate. To avoid such error, we can prepare a data-logger system connected with computer to record the temperature change of the solution.
Secondary, there must be some heat loss to the surroundings. It will cause the enthalpy change of the solution smaller than the expected. Then the molar enthalpy change of the solutions also decrease, as a result, the enthalpy change of hydration of calcium carbonate will be smaller than the expected. To get rid of the heat to surroundings, we use the vacuum flask instead of the polystyrene cup to prevent heat loss by radiation. Cover the cup by a plastic lid to prevent heat loss by convection. Moreover, we can place the cup on the heat-insulating plate in order to reduce hest loss by conduction.
On top of that, we had used the measuring cylinder to measure 25 cm3 of acid or alkali which is less precise than using the pipette to get 25cm3 of acid or alkali. As the volume we got was not equal to exact 25 cm3 of acid or alkali, the experimental enthalpy change of neutralization is smaller or greater than the expected value.
Questions
Q.1:
The temperature increased in neutralization between 25 cm3 of 2.0 M of HCl
and 25 cm3 of 2.0 M of NaOH was 12.0 . The temperature increased in neutralization between 25 cm3 of 2.0 M HCl and 25 cm3 of 4.0M of NaOH was 13.5 .
Q.2:
The energy released in neutralization between 25 cm3 of 2.0 M of HCl
and 25 cm3 of 2.0 M of NaOH was 2567J. The energy released in neutralization
between 25 cm3 of 2.0 M HCl and 25 cm3 of 4.0M of NaOH was 2889J.
Q.3:
The no of mole of water formed in neutralization between 25 cm3 of 2.0 M of HCl and 25 cm3 of 2.0 M of NaOH was 0.05mol. The no of mole of water formed in neutralization between 25 cm3 of 2.0 M HCl and 25 cm3 of 4.0M of NaOH was 0.1 mol.
Q.4:
The enthalpy change of neutralization between 25 cm3 of 2.0M of HCl and 25 cm3 of 2.0M of NaOH is –51.3KJ mol-1. The enthalpy change of neutralization between 25 cm3 of 2.0M of HCl and 25 cm3 of 4.0M of NaOH is –28.9KJ mol-1.
Q.5:
H+ (aq) + OH- (aq) → H2O (l)
Q.6:
By the detailed calculation in the Result and Calculation section, the enthalpy change of neutralization is –54.8KJ mol-1
Q7:
H+ (aq) + OH- (aq) → H2O (l)
Q8:
By the detailed calculations in the result and calculation section, the enthalpy change of neutralization is –44.9KJ mol-1
Q9:
The enthalpy change of neutralization of nitric acid and sodium hydroxide is the smallest while the enthalpy change of neutralization of hydrochloric acid and potassium hydroxide is the greatest. Their values of enthalpy change of neutralization are high because strong acids and bases are used in this experiment. They can completely dissociate in the water, so the no of mole of H+ and OH- in the solution increases; as a result, the enthalpy changes of neutralization are relatively higher. In addition, there is no energy that is needed for the ionization of the hydrogen and hydroxide ions. However, nitric acid used in the experiment is very diluted and it is slightly less strong than hydrochloric acid. Therefore, the enthalpy change of neutralization of nitric acid is relatively lower than that of hydrochloric acid.
Q10:
Both accepted values of the first two reactions are greater than that of the experimental values due to several experimental errors, which is discussed above in the source of errors section.
Q11:
No. It is because the enthalpy change of neutralization of 1st reaction is greatest while the enthalpy change of neutralization of 3rd reaction is smallest. It is different from the experiments. In addition, hydrogen bromide is used in 2nd reaction while it is not used in our experiment. Hydrogen bromide is weaker than hydrochloric acid. So it does not agree with explanation in Q.9.
Q12:
As the dissociation constants of hydrochloric acid, nitric acid and sulphric acid are much larger than that of ethanoic acid, it means ethanoic acid is a acid which is much weaker than them. As it is a weak acid, it only partially dissociate in the solution.
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H+ (aq)
Therefore, a part of energy from neutralization has lost tp push the reaction towards the right side. It results in the enthalpy change of neutralization of ethanoic acid is smaller than that of the strong acid.
H2S (aq) + H2O (l) HS- (aq) + H3O+ (aq)
In fact, there is no hydrogen ion formed in the solution of the process of dissociation, so energy is needed for has lost to push the reaction towards right side and the formation of hydrogen ions. Furthermore, the dissociation constant of hydrogen sulphide is even lower than the ethanoic acid which means hydrogen sulphide is weaker than ethanoic acid. As a result, the enthalpy change of neutralization of hydrogen sulphide is substantially lower.
Q13:
The temperature increased of neutralization between 2.0M of sulphuric acid and 2.0Mof sodium hydroxide is much lower than that between 2.0M of sulphuric acid and 4.0M of sodium hydroxide due to the concentration effect. From the equation, 1 mole of sulphuric acid reacts with 2 mole of sodium hydroxide. So the neutralization between 2.0M of sulphuric acid and 2.0Mof sodium hydroxide is an incomplete reaction because the mole ratio is 1: 1. While neutralization between 2.0M of sulphuric acid and 4.0M of sodium hydroxide is a complete reaction because the mole ratio is 1: 2. As a result the temperature increased of neutralization between 2.0M of sulphuric acid and 4.0M of sodium hydroxide is much higher than that of neutralization between 2.0M of sulphuric acid and 2.0Mof sodium hydroxide.
Q14
The temperature increased of neutralization between 2.0M of hydrochloric acid and 2.0Mof sodium hydroxide is much lower than that between 2.0M of sulphuric acid and 4.0M of sodium hydroxide due to the difference of no of mole of hydrogen dissociated in the solution. As sulphuric acid is dibasic acid while hydrochloric acid is monobasic acid. As a result, the amount of hydrogen ionized by the sulphuric acid is higher than that of hydrochloric acid. Furthermore, from the equation, 1 mole of hydrochloric acid reacts with 1 mole of sodium hydroxide. So the neutralization between 2.0M of hydrochloric acid and 4.0M of sodium hydroxide is an incomplete reaction because the mole ratio is 1: 2. While neutralization between 2.0M of sulphuric acid and 4.0M of sodium hydroxide is a complete reaction because the mole ratio is 1: 1. As a result the temperature increased of neutralization between 2.0M of sulphuric acid and 4.0M of sodium hydroxide is much higher than that of neutralization between 2.0M of hydrochloric acid and 2.0Mof sodium hydroxide.
Conclusion
From the above six neutralization, we can calculate the enthalpy change of neutralization by (m1c1 + m2c2) * Temp. Difference. Through the comparisons between the reaction, we can know about how the effect of concentration, basicity and straightness of acid and base affect the enthalpy change of neutralization.