Calculation for measuring enthalpy changes

[Q = m∙c∙∆T]

Q = (50.00g ± 1.0g) x (4.18J∙g-1∙k-1) x (-13) = -2717J

Conversion of Uncertainty to Relative Error

[(Nearest unit of volume measurement observed ÷ desired volume) x 100 = Percentage error of volume]

(1÷50) x100= 2%

[(Nearest unit of temperature measurement observed ÷ recorded temperature)

x 100 = Percentage error of recorded temperature]

(1÷13) x 100= 7.7%

[Percentage value of volume error + percentage value of temperature error = relative error]

2%+7.7% = 9.7% relative error

Enthalpy change in joules with relative error percentage

Q= -2717J ± 9.7%

Conversion of Relative Error to Absolute Error

(2717÷100) x 7.7= 209 Absolute Uncertainty = ±209J → Q= -2717J ± 209J

(÷1000) [To change to kilojoules]

Q= -2.72 kJ ± 0.209kJ (÷0.050mol H2O) [To find kilojoules per mole of water]

Q =-54.3 kJ∙mol-1 H2O (± 4.18 kJ∙mol-1 H2O)

Calculating Percentage Error of Observed Enthalpy Value

[(Observed Value – Expected Value) ÷ Expected Value] x 100 = Percentage error

[(-54.3 kJ∙mol-1 H2O (± 4.18 kJ∙mol-1 H2O) – 57.9 kJ∙mol-1) ÷ 57.9 kJ∙mol-1] x 100 = 6.21%

Experiment 2 – Enthalpy of neutralization for hydrochloric acid and potassium hydroxide

KOH(aq) + HCl(aq)→KCl(aq) + H2O(l)

Finding the Moles of Water

Moles of hydrochloric acid:

[Volume x Concentration]

0.025dm3 x 2.00mol∙dm-3 = 0.050mol HCl

Moles of potassium hydroxide:

[Volume x Concentration]

0.025dm3 x 2.00mol∙dm-3 = 0.050mol KOH

Moles of water:

[Equal to moles of limiting reagent]

0.050mol H2O

Calculating Enthalpy of Neutralization

Mass/volume of solution

[Mass = Volume, if we assume 1g = 1cm3]

25.00cm3 (± 0.5cm3) + 25.00cm3 (± 0.5cm3) = 50.00cm3 (± 1.0cm3) = 50.00g (± 1.0g)

Calculation for measuring enthalpy changes

[Q = m∙c∙∆T]

Q = (50.00g ± 1.0g) x (4.18J∙g-1∙k-1) x (-12) = -2508J

Conversion of Uncertainty to Relative Error

[(Nearest unit of volume measurement observed ÷ desired volume) x 100 = Percentage error of volume]

(1÷50) x 100= 2%

[(Nearest unit of temperature measurement observed ÷ recorded temperature)

x 100 = Percentage error of recorded temperature]

(1÷12)x100= 8.3%

[Percentage value of volume error + percentage value of temperature error = relative error]

2%+8.3% = 10.3% relative error

Enthalpy change in joules with relative error percentage

Q= -2508J ± 10.3%

Conversion of Relative Error to Absolute Error

(2508÷100) x 10.3= 258 Absolute Uncertainty = ±258J → Q= -2508J ± 258J

(÷1000) [To change to kilojoules]

Q= -2.51 kJ ± 0.258kJ (÷0.050mol H2O) [To find kilojoules per mole of water]

Q = -50.2 kJ∙mol-1 H2O (± 5.17 kJ∙mol-1 H2O)

Calculating Percentage Error of Observed Enthalpy Value

[(Observed Value – Expected Value) ÷ Expected Value] x 100 = Percentage error

[(-50.2 kJ∙mol-1 H2O (± 5.17 kJ∙mol-1 H2O) – -57.3 kJ∙mol-1) ÷ 57.3 kJ∙mol-1] x 100 = 12.4%

Experiment 3 – Enthalpy of neutralization for nitric acid and sodium hydroxide

NaOH(aq) + HNO3(aq)→ NaNO3(aq) + H2O(l)

Finding the Moles of Water

Moles of nitric acid:

[Volume x Concentration]

0.025dm3 x 2.00mol∙dm-3 = 0.050mol HCl

Moles of sodium hydroxide:

[Volume x Concentration]

0.025dm3 x 2.00mol∙dm-3 = 0.050mol KOH

Moles of water:

[Equal to moles of limiting reagent]

0.050mol H2O

Calculating Enthalpy of Neutralization

Mass/volume of solution

[Mass = Volume, if we assume 1g = 1cm3]

25.00cm3 (± 0.5cm3) + 25.00cm3 (± 0.5cm3) = 50.00cm3 (± 1.0cm3) = 50.00g (± 1.0g)

Calculation for measuring enthalpy changes

[Q = m∙c∙∆T]

Q = (50.00g ± 1.0g) x (4.18J∙g-1∙k-1) x (-12.75) = -2665J

(-12.5 + -13) ÷ 2

Conversion of Uncertainty to Relative Error

[(Nearest unit of volume measurement observed ÷ desired volume) x 100 = Percentage error of volume]

(1÷50) x 100= 2%

[(Nearest unit of temperature measurement observed ÷ recorded temperature)

x 100 = Percentage error of recorded temperature]

(1÷12.75)x100= 7.8%

[Percentage value of volume error + percentage value of temperature error = relative error]

2%+7.8% = 9.8% relative error

Enthalpy change in joules with relative error percentage

Q= -2665J ± 9.8%

Conversion of Relative Error to Absolute Error

(2665÷100) x 9.8= 261 Absolute Uncertainty = ±261J → Q= -2665J ± 261J

(÷1000) [To change to kilojoules]

Q= -2.67 kJ ± 0.261kJ (÷0.050mol H2O) [To find kilojoules per mole of water]

Q = -53.4 kJ∙mol-1 H2O (± 5.22 kJ∙mol-1 H2O)

Calculating Percentage Error of Observed Enthalpy Value

[(Observed Value – Expected Value) ÷ Expected Value] x 100 = Percentage error

[(-53.4 kJ∙mol-1 H2O (± 5.22 kJ∙mol-1 H2O) – -57.6 kJ∙mol-1) ÷ 57.6 kJ∙mol-1] x 100 = 7.47%

Experiment 4.1 –

Enthalpy of neutralization for sulfuric acid and 2.00mol∙dm-3 sodium hydroxide

2NaOH(aq) + H2SO4(aq)→ Na2SO4(aq) + 2H2O(l)

Finding the Moles of Water

Moles of sulfuric acid:

[Volume x Concentration]

0.025dm3 x 2.00mol∙dm-3 = 0.050mol H2SO4

Moles of sodium hydroxide:

[Volume x Concentration]

0.025dm3 x 2.00mol∙dm-3 (x 2) = 0.100mol NaOH

Moles of water:

[Equal to moles of limiting reagent]

0.050 H2SO4 → 0.050mol H2O (x2) = 0.100mol H2O

Calculating Enthalpy of Neutralization

Mass/volume of solution

[Mass = Volume, if we assume 1g = 1cm3]

25.00cm3 (± 0.5cm3) + 25.00cm3 (± 0.5cm3) = 50.00cm3 (± 1.0cm3) = 50.00g (± 1.0g)

Calculation for measuring enthalpy changes

[Q = m∙c∙∆T]

Q = (50.00g ± 1.0g) x (4.18J∙g-1∙k-1) x (-12.25) = -2560J

(-12 + -12.5) ÷ 2

Conversion of Uncertainty to Relative Error

[(Nearest unit of volume measurement observed ÷ desired volume) x 100 = Percentage error of volume]

(1÷50) x 100= 2%

[(Nearest unit of temperature measurement observed ÷ recorded temperature)

x 100 = Percentage error of recorded temperature]

(1÷12.25)x100= 8.2%

[Percentage value of volume error + percentage value of temperature error = relative error]

2%+8.2% = 10.2% relative error

Enthalpy change in joules with relative error percentage

Q= -2560J ± 10.2%

Conversion of Relative Error to Absolute Error

(2560÷100) x 10.2= 261 Absolute Uncertainty = ±261J → Q= -2560J ± 261J

(÷1000) [To change to kilojoules]

Q= -2.56 kJ ± 0.261kJ (÷0.050mol H2O) [To find kilojoules per mole of water]

Q = -51.2 kJ∙mol-1 H2O (± 5.22 kJ∙mol-1 H2O)

Calculating Percentage Error of Observed Enthalpy Value

[(Observed Value – Expected Value) ÷ Expected Value] x 100 = Percentage error

[(-51.2 kJ∙mol-1 H2O (± 5.22 kJ∙mol-1 H2O) – -57.3 kJ∙mol-1) ÷ 57.3 kJ∙mol-1] x 100 = 10.6%

Experiment 4.2 –

Enthalpy of neutralization for sulfuric acid and4.00mol∙dm-3 sodium hydroxide

2NaOH(aq) + H2SO4(aq)→ Na2SO4(aq) + 2H2O(l)

Finding the Moles of Water

Moles of sulfuric acid:

[Volume x Concentration]

0.025dm3 x 2.00mol∙dm-3 = 0.050mol H2SO4

Moles of sodium hydroxide:

[Volume x Concentration]

0.025dm3 x 4.00mol∙dm-3 = 0.100mol NaOH (x2) = 0.200mol NaOH

Moles of water:

[Equal to moles of limiting reagent]

0.050mol H2SO4 → 0.050mol H2O(x2) = 0.100mol H2O

Calculating Enthalpy of Neutralization

Mass/volume of solution

[Mass = Volume, if we assume 1g = 1cm3]

25.00cm3 (± 0.5cm3) + 25.00cm3 (± 0.5cm3) = 50.00cm3 (± 1.0cm3) = 50.00g (± 1.0g)

Calculation for measuring enthalpy changes

[Q = m∙c∙∆T]

Q = (50.00g ± 1.0g) x (4.18J∙g-1∙k-1) x (-14) = -2926J

Conversion of Uncertainty to Relative Error

[(Nearest unit of volume measurement observed ÷ desired volume) x 100 = Percentage error of volume]

(1÷50) x 100= 2%

[(Nearest unit of temperature measurement observed ÷ recorded temperature)

x 100 = Percentage error of recorded temperature]

(1÷14)x100= 7.1%

[Percentage value of volume error + percentage value of temperature error = relative error]

2%+7.1% = 9.1% relative error

Enthalpy change in joules with relative error percentage

Q= -2926J ± 9.1%

Conversion of Relative Error to Absolute Error

(2926÷100) x 9.1= 266 Absolute Uncertainty = ±266J → Q= -2926J ± 266J

(÷1000) [To change to kilojoules]

Q= -2.93 kJ ± 0.266kJ (÷0.050mol H2O) [To find kilojoules per mole of water]

Q = -58.5 kJ∙mol-1 H2O (± 5.32 kJ∙mol-1 H2O)

Calculating Percentage Error of Observed Enthalpy Value

[(Observed Value – Expected Value) ÷ Expected Value] x 100 = Percentage error

[(-58.5 kJ∙mol-1 H2O (± 5.32 kJ∙mol-1 H2O) – -57.3 kJ∙mol-1) ÷ 57.3 kJ∙mol-1] x 100 = 2.09%

Table 5.1: Accepted values for enthalpy of neutralization vs. observed values

1: I looked all over the internet and couldn’t find a solid answer or source to take an accepted value of enthalpy of neutralization for these reactions. The sources answers I found ranged from -55.9kJ∙mol-1 to -59.80kJ∙mol-1. Wikipedia’s ‘Enthalpy of Neutralization’ page strangely assumes that the enthalpy change of neutralization for “a strong acid and base” is -57.3kJ∙mol-1.

“The standard enthalpy change of neutralization for a strong acid and base is -57.3 kJ/mol”

Taken from

So because this was the only page to have given an answer, though likely invalid, it was more credible than the majority of other sites where I could have drawn information from, which included forums, yahoo answers, blogs, and one-off question sites.

Table 6.1: Error Propagation Table

Conclusion

Our experiments each yielded results that were less than 10% in error to the accepted values of enthalpy of neutralization for our 5 reactions. In the table above is our percentage error value for each experiment’s enthalpy change. This could be due to a number of reasons, including our equipment’s precision, some uncontrolled variables (room temperature, human contact with the container) and random error such as misreading of the volume of solutions or thermometer’s display ( caused by parallax).

Our equipment throughout the experiment was basic in nature, made for a high school chemistry laboratory and was not as precise as it could have been if we had been aiming for prime results. Our volumes had an uncertainty of plus/minus 0.5ml which gave us a constant 2% absolute error for each measurement of volume, while our thermometer had an uncertainty of plus/minus 0.5°C, giving us a varying absolute error value depending on our recorded result.

Evaluation

As stated in our conclusion, there were many possible sources our errors throughout the experiment could have come from. Systematic errors include the degree of precision for our instruments and materials, and we could improve the experiment’s accuracy by having simply more precise equipment such as measuring beakers with a 0.05ml uncertainty rather than 0.5ml, or a data logger to measure temperature rises rather than having students observe and measure, making the reading susceptible to random errors such as parallax and simply human error.

Another factor not taken into account was the room temperature, though kept more or less even, it was not constant, and therefore could have affected the experiment’s results if it had changed the temperature of the reagents or final solution by even one degree. To improve on this possible error, we could attempt the experiment in a space where the temperature is constant. Seeing as temperature would be a prime affected factor in the experiment we also have to take into account the fact that by simply touching the materials (measuring beakers, thermometers etc.) that we may also be passing heat on through means of conduction by our own bodies/hands. So to improve upon this, we could have attempted the experiment wearing insulating gloves to reduce heat given off by ourselves to the materials.