Determinaing Enthalpy Change of Potassium

Topics: Thermodynamics, Enthalpy, Energy Pages: 7 (2164 words) Published: December 19, 2010
Determining the enthalpy change for the thermal decomposition of potassium hydrogen carbonate into potassium carbonate. Controlled Variables:
1. Volume of HCl ± 0.5 cm3 (± 2%)
2. Concentration of HCl,
3. Same mass of K2CO3 and KHCO3 within specified ranges of 2.5 – 3.0g and 3.25 – 3.75g respectively 4. Same calorimeter used i.e. polystyrene cup is used in this experiment 5. Same thermometer will be used ± 0.10K

6. Same source of K2CO3, KHCO3 and HCl

Raw Data Results:
The raw data table shows, the temperatures at initial point, after and the change in temperature of the reaction between K2CO3(s) with HCl (aq) The change in temperature is calculated as:
After temperature – initial temperature= change in temperature K2CO3| The temperatures of each trial at ± 0.10K|
Trials| Initial ± 0.5%| After ± 0.4%| Change ± 0.9%| 1| 21.600| 23.700| -2.100|
2| 22.200| 24.600| -2.400|

Therefore the Average can be calculated for the temperature: The equation shows, Average = change in temperature of trials (1 + 2)/2 Therefore the Average = (-2.1 +( - 2.4))/2
= -2.250K ± 0.9%
The table below shows the raw data of the temperatures at their initial point, after and the change in temperature of the reaction between KHCO3 with HCl. KHCO3| The temperatures of each trial at ± 0.10K|

Trials| Initial ± 0.5%| After ± 0.7%| Change is ( + ) ± 1.2%| 1| 21.700| 14.300| 7.4|
2| 22.000| 14.200| 7.8|

The average can be calculated for the temperature by using the same equation as shown above. Therefore: Average of the temperature change of KHCO3 = (7.4 + 7.8)/2
= 7.6 0K ± 1.2%
When using a thermometer that measures ± 0.1 K the uncertainty shown in the table of trial 1 for example, K2CO3 measures 21.7000K will equal to (0.1/21.600) X 100 = ± 0.5%
for trial 2:
(0.1/14.300) X 100 = ± 0.4 %
The overall uncertainty for K2CO3 for the temperature change is the summing up of all the uncertainties. 0.5 + 0.4 = ± 0.9% ( this has been used above )
This is the same method for the calculation of the uncertainties for KHCO3, For trial 1:
(0.1/21.000) X 100 = ± 0.5 %
For trial 2:
(0.1/ 21.700) X 100 = ± 0.7 %
The overall for KHCO3 is
0.5 + 0.7 = ± 1.2 %

As the mass was used from a range, for K2CO3 and KHCO3 from specified ranges of 2.5 – 3.0g and 3.25 – 3.75g respectively. Therefore: The raw data for the two trials for both the compounds has been tabled with the average worked out for each species.

For K2CO3| For KHCO3|
Trials| The mass that was used from specified range of 2.5 – 3.0g ± 0.010g| Average| Trials| The mass that was used from specified range of 3.25 – 3.75g ± 0.010g| Average| 1| ± 0.4%2.729± 0.4%2.613| ± 0.8%| 1| ± 0.3%| ± 0.6%| | | 2.671| | 3.455| 3.555|

2| | | 2| ± 0.3%| |
| | | | 3.665| |

The average was worked out as:
= the mass of trials (1 + 2)/2
When using a balance that weighs ± 0.010g the uncertainty shown in the table of trial 1 for K2CO3 weighs 2.729g will equal to (0.010/2.729) X 100 = 0.4%
Hence for trial 2 =
(0.010/2.613) X 100 = 0.4%
Therefore the overall uncertainty for the mass of K2CO3 is summing up all the individual uncertainties for both the trials. Hence 0.4 + 0.4
= 0.8%
For KHCO3:
Trial 1:
(0.010/3.455) X 100 = 0.3%
Trial 2:
(0.010/3.665) X 100 = 0.3%
Therefore the overall uncertainty equals
0.3 + 0.3 = 0.6%
The moles of each species are required to be calculated. The calculation is, Moles = mass/Mr
Therefore the Mr of K2CO3 = (2 X K) + (1 X C) + (3 X 0)
Mr = (2 X 39.10) + 12.01 + (3 X 16)
= 138.01
Therefore as the mass is 2.671, then
Moles = 2.67/ 138.21
= 0.0193 M
For KHCO3
Mr = 39.10 + 12.01 + (3 X 16) + 1.01
= 100.11 ± 4 %
Moles = 3.555/ 100.11
= 0.0354 ± 3 %
Specific heat capacity of solution (Jg-1 0C-1)
Temp. change (0C)
Mass of solution being heated (g)
The enthalpy change is required to be calculated....

Please join StudyMode to read the full document