Chemistry Lab; the neutralization of NaOH and Hess Law

Research question:

Is the reaction endothermic or exothermic and can Hess law be used to calculate the enthalpy change?

Introduction

The aim of the lab is to find out if the neutralization of NaOH and HCl is exothermic or endothermic, and if we can calculate the enthalpy change of the reaction c using Hess law.

• The independent variable is the amount of substance and the actual substance used in the reaction.
• The dependent variable is the temperature change of the reactions
• The controlled variable is the time recorded for each reaction.

Equipment:

Digital thermometer ± 0.1°, watch, magnetic stirrer, 100 ml PE beakers, tripod and clamp, burette ± 0.04 ml, scale ±0.01g.

Chemicals:

Sodium hydroxide (solid), solutions of sodium hydroxide and hydrochloric acid with concentration 1,00 moldm^-³.¨

Procedure:

Reaction a:

1. 50 cm3 of water was measured up and transferred into a plastic beaker on a magnetic stirrer. The temperature was recorded.
2. 1.0 g of NaOH(s) was added. The mass of the pastilles varied slightly. The desired value was reached.
3. The pastilles were then dissolved during continues stirring. The temperature of the reaction was then measured every 30 seconds under a time period of 240 seconds.

Reaction b:

1. 25 cm3 of 1.0 moldm^-3 hydrochloric acid was measured up. The solution was placed on the stirrer and the temperature was recorded once again.

2. The same volume of 1.0 moldm^-3 sodium hydroxide solution was added.

3. The temperature of the reaction was then measured every 30 seconds under a time period of 240 seconds.

Reaction c:

1. 25 cm3 of 1.0 moldm^-3 hydrochloric acid was measured up + 25cm3 deionised water. The solution was placed on the stirrer and the temperature was recorded once again.
2. 1.0 grams of NaOH was added.
3. The pastilles were then dissolved during continues stirring. The temperature of the reaction was then measured every 30 seconds under a time period of 240 seconds.

H2O

Reaction A: NaOH(s)àNa⁺_(aq) + OH^-_(aq)

Table A: Temperature of the reaction measured every 30 seconds:

Time(sec)	Temperature (°c)
0	20.3
30	21.5
60	22.3
90	22.7
120	23.2
150	23.6
180	23.8
210	23.9
240	24.1

Graph A

Temperature change during the reaction:

Calculation of Enthalpy Change (ΔH) during the reaction A:

Mass (m) x Specific heat capacity (c) x temperature change (ΔT) = Heat energy

ΔT = 24.1-20.3 = 2.8 K

50cm3 = 50g = 0.05 kg

Heat energy = 0.05 kg x 4.18 c x 3.8 K = 0.7942 kJ

Amount of NaOH = 1/40 = 0.025

As the reaction is exothermic, the enthalpy change (ΔH) will be:

-0.7942 x (1/0.025) = -31.768 kJ mol^-1

Reaction B: Na⁺ + OH^- + Cl à Na⁺ + Cl^- +H2O

Table B: Temperature change of the substance every 30 seconds:

Time (s)	Temperature (°c)
0	20.2
30	26.0
60	25.9
90	25.8
120	25.7
150	25.6
180	25.5
210	25.4
240	25.3

Graph B

Temperature change of the substance every 30 seconds:

Calculation of Enthalpy Change (ΔH) during the reaction B:

Mass (m) x Specific heat capacity (c) x temperature change (ΔT) = Heat energy

ΔT = 26-20.2 = 5.8 K

50cm3 = 50g = 0.05 kg

Heat energy = 0.05 kg x 4.18 c x 5.8K = 1.2122 kJ

Amount of HCL = 25/1000 = 0.025 mol

Amount of NaOH = 25/1000 = 0.025 mol

As the reaction is exothermic, the enthalpy change (ΔH) will be:

– 1.2122 x (1/0.025) = -48.488 kJ mol^-1

H2O

Reaction C: NaOH+H⁺+Cl^- à Na⁺ +Cl^-+H₂O

Table C: Temperature change of the substance every 30 seconds:

Time (sec)	Temperature (°c)
0	19.9
30	22.6
60	24.7
90	26.9
120	28.0
150	29.3
180	29.8
210	30.0
240	30.0

Graph C

Temperature change of the substance every 30 seconds:

Calculation of Enthalpy Change (ΔH) during the reaction C:

Mass (m) x Specific heat capacity (c) x temperature change (ΔT) = Heat energy

ΔT = 30-20 = 10 K

50cm3 = 50g = 0.05 kg solution

Heat energy = 0.05 kg x 4.18 c x 10 K = 2.09 kJ

Amount of HCL = 25/1000 = 0.025 mol

As the reaction is exothermic, the enthalpy change (ΔH) will be:

– 2.09 x (1/0.025) = -83.6 kJ mol^-1

Calculating the enthalpy change (ΔH) of Reaction 3 Using Hess Law:

ΔH₁ or ΔH of reaction A = -31.768

ΔH₂ or ΔH of reaction B = -48.488

ΔH₃ or ΔH of reaction C = X

Using Hess Law we can now calculate the enthalpy change (ΔH) of ΔH₃ = X

ΔH₁ + ΔH₂ = ΔH₃

-31.768 + -48.488 = -80.256 kJ mol ^-1 = ΔH₃

Conclusion

From graphs a, b and c you can state that all reactions were exothermic, as the reaction takes place temperature rises as time passes. In graph b the curve declines after 30 seconds, this is because the reaction takes place in the first 30 seconds (with a temperature rise of 20 degrees Celsius), and then the substance begins to cool down as the reaction has come to an end.

The enthalpy change of reactions a, b and c have been calculated, then the enthalpy change of reaction c has been calculated using Hess Law in comparison to the former calculation when the data from the experiment was used. As can be seen the results are somewhat similar and correct:

· The result of the calculation of reaction c: -83.6 kJ mol ^-1, compared to;

· The enthalpy change calculated using Hess Law: -80.256 kJ mol ^-1

The slight difference between these two values is due to errors during measurements of the reactions temperature, weight of substance, volume, time etc. For example the thermal thermometer, the burette and the scale which was used are not entirely precise as described earlier in the lab, so one way of getting a more accurate result would be using more precise equipment to measure different aspects of the reactions.

If we look at the preciseness of using Hess Law to calculate the enthalpy of reaction c, we can state that it can be used to give the exact enthalpy change of the reaction, given that the calculated values needed for the use of Hess Law are correct (calculated values of enthalpy change for reactions c and b).

The improvements that could be made for a more accurate result could be as earlier mentioned more precise equipment, but also the temperature of the substances initially were not measured, which can have affected the rate of which the different reactions took place. If a double boiler would have been used, the reactions would have taken place at the same initial temperature making the experiment and collected data more accurate.

The reactions which took place were all exothermic, and Hess Law could be used to calculate the enthalpy change, although it would have been more precise given that the raw data was more accurate.