ΔHr
2NaHCO3 (s) Na2CO3(s) + H2O(l) + CO2(g)
ΔH1 2HCl 2HCl ΔH2
2NaCl(aq) + H2O(l) +CO2(g)
As an alternative to directly calculating the enthalpy change during the decomposition of sodium hydrogen carbonate, represented above as ΔHr, a different approach is required and according to Hess’s Law, ΔHr could be calculated by following a different route, in this case calculating the enthalpy changes that occur during the reactions of ΔH1 (NaHCO3(s) + HCl(aq)) and ΔH2 (Na2CO3(s) + 2HCl(aq)) would allow me to evaluate a reasonable estimate to the enthalpy of ΔHr using the formula below.
ΔHr = ΔH1 - ΔH2
Apparatus
-
50cm3 cylinder for measuring out HCl solution
-
A 250 cm3 bottle of 2 Molar solution Hydrochloric Acid
- A container of Sodium Hydrogen Carbonate Powder
- A container of Sodium Carbonate Powder
- A polystyrene cup for insulation against heat loss or gain during reaction
- Spatula for measuring out the required amount of powder
- Digital scales (to the nearest hundredth of a gram) for weighing powder
- Paper tray to prevent loss of powder when measuring on scales
- Thermometer for estimating temperatures
- Stirring Rod to stir dissolving powders fully
- Safety goggles
- Pencil and paper to write results
It is necessary to calculate the amounts of powder needed in this experiment.
2NaHCO3(s) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g)
Na2CO3(s) + 2HCl(aq) 2NaCl(aq) + H2O(l) +CO2(g)
It is possible to calculate the amount in grams of each powder of either sodium hydrogen carbonate or sodium carbonate required to fully react with 50cm3 of HCl using the above equations.
The number of moles in 50cm3 of 2 molar acid is (50/1000) x2 = 0.1 moles.
The R.A.M of 1 mole of (NaHCO3) is (23+1+12+48) = 84
1 mole of (NaHCO3) reacts with 1 mole of (HCl) and so the number of grams of (NaHCO3) required is 84 x 0.1 = 8.4 grams
The R.A.M of 1 mole of (Na2CO3) is (46+12+48) = 106
1 mole of (Na2CO3) reacts with 2 moles of (HCl) and so the number of grams of (Na2CO3) required is 106 x (0.1/2) = 5.3 grams
The experiment will involve 8.4 grams of (NaHCO3) and 5.3 grams of (Na2CO3) but to allow for a complete reaction and any errors in measurements or loss of powder I will add one extra gram to each of the substances to ensure a better result.
The measurements for the experiment are:
- 9.4 grams of Sodium Hydrogen Carbonate
- 6.3 grams of Sodium Carbonate
Method
Prepare the list of apparatus and wear safety goggles throughout to prevent hazardous substances like HCl entering eyes. Measure out 50cm3 of HCl solution using the measuring cylinder and the bottle of 2 molar solution, remembering to judge the content by viewing the meniscus at eye level. Pour the HCl directly into the polystyrene cup avoiding spillages and ensuring that as little acid remains in the cylinder. Leave the acid in the cup and using the thermometer measure the ambient temperature of the acid and record the result. Use the spatula, the paper tray, the digital scales and the container of either the sodium hydrogen carbonate to weigh out 9.4 grams or the sodium carbonate to weigh out 6.3 grams. When weighing out correct amounts make sure not to spill any of the contents and disrupt the results, also check that the scales you are using have been set to zero before weighing and that any contaminants have been removed that could effect measurements. After weighing out the powder re-check the temperature of the acid in the polystyrene cup and alter results if necessary. With the thermometer remaining in the cup add the powder and stir with either the stirring rod or thermometer to improve the chances of a complete reaction. During the reaction the thermometer will either show a temperature increase or decrease, record the maximum temperature change along with the appropriate powder name and mass. Repeat this experiment three times for both sodium hydrogen carbonate and sodium carbonate powders recording the temperature changes before the powder is added to the solution and the maximum temperature change after the powder is added.
Results
To calculate the enthalpy I must assume that 1cm3 of HCl is equivalent to 1 gram of H2O because I must use the specific heat of water, which is the amount of energy required to raise one gram of water by one degree Celsius (4.18 Joules) in the formula:
Enthalpy change = mass or volume of a substance x 4.18 Joules x change in temperature in degrees Celsius
Calculations for ΔH1:
Mass of NaHCO3 used = 9.43 grams, 9.41 grams and 9.44 grams
Volume of HCl (2M) used = 50cm3
For ΔH1 the temperature changes in order were 8°C, 10°C and 10.5°C
Heat taken in when NaHCO3 reacts with HCl =
- 50 x 4.18 x 8 = 1672 Joules
- 50 x 4.18 x 10 = 2090 Joules
- 50 x 4.18 x 10.5 = 2194.5 Joules
Moles of NaHCO3 used by 50cm3 of 2 Molar HCl = (50/1000) x2 = 0.1 Moles
The average enthalpy change for 0.1 Moles of NaHCO3 = (1672+2090+2194.5)/3 = 1985.5 Joules
For 1 mole of NaHCO3 = 1985.5 Joules x 10 = 19,855 Joules
Calculations for ΔH2:
Mass of Na2CO3 used = 6.32 grams, 6.38 grams and 6.33 grams
Volume of HCl (2M) used = 50cm3
For ΔH2 the temperature changes in order were 0.5°C, 1.0°C and 2.0°C
Heat given out when Na2CO3 reacts with HCl =
- 50 x 4.18 x 0.5 = 104.5 Joules
- 50 x 4.18 x 1.0 = 209 Joules
- 50 x 4.18 x 2.0 = 418 Joules
Moles of Na2CO3 used by 50cm3 of 2 Molar HCl = ((50/1000) x2)/2 = 0.05 Moles
1 mole of Na2CO3 would produce twenty time more energy = 0.05 x 20 = 1 mole
The average enthalpy change for 0.05 Moles of Na2CO3 = (104.5+209+418)/3 = 243.84 Joules
For 1 mole of Na2CO3 = 243.84 Joules x 20 = 4876.7 Joules
I have two results now, ΔH1 is 19,855 Joules and ΔH2 is 4876.7 Joules to substitute them into the equation I have to write it like this:
ΔHr = ΔH1 - ΔH2
ΔHr = 19,855 Joules – -4876.7 Joules
ΔHr = 24731.7 Joules
Conclusion
I can now conclude that from my results that the enthalpy change for the decomposition of sodium hydrogen carbonate when heated is approximately 24731.7 Joules.
To perform this experiment I have had to make several assumptions, for example, I have had to assume that the specific heat of HCl is the same as that of water, ‘to raise the temperature of a gram of water by one degree’s Celsius would require 4.18 Joules of energy’. I have also assumed that the solution of acid I am using is exactly 2 Molar and that throughout my experiments I have had no loss in liquids or powders due to human fault.
To improve the experiment I could have made my own acid to 2 molar to increase the accuracy of my results. I could have used more sophisticated equipment, such as a more accurate digital scale or thermometer. A more appropriate lab with set conditions and both constant temperature and pressure would be more advisable than a school classroom where conditions vary randomly.