I then accurately weighed the plastic weighing bottle on scales accurate to three decimal places. The weight came to 6.136g. I then added 3.0g of dry sodium hydrogen carbonate, ensuring that the weight of NaHCO3 was slightly over the stated 3.0g. This will help to minimise error, as some of the NaHCO3 might be left in the weighing bottle when we add it to the acid, or could stick to the sides of the calorimeter, whereas we want to ensure that all of the stated 3.0g of NaHCO3 react with the acid. The weight of the bottle plus the NaHCO3 was 9.339g.
I then added the dry NaHCO3 to the 50 cm3 of 2M HCl in the polystyrene cup, and quickly stirred it to ensure the reactants were completely homogenised. I then recorded the temperature of the solution at 30 s intervals as it fell steadily, until the temperature stabilised. This took about ten minutes. I added these results to the aforementioned table (please see separate sheet).
I weighed the empty weighing bottle once again, (having emptied the NaHCO3 into the calorimeter), and the weight was 6.136g. I then deducted the result of this from the second weight (bottle + 3.0g NaHCO3): this will give us a more accurate figure for the actual mass of NaHCO3 used, as it accounts for possible error in any material left in the weighing bottle. I found that the actual mass of NaHCO3 I used was 3.203g.
After having collected all the results in my table, I used them to plot a graph of temperature against time for this reaction between NaHCO3 and HCl. I then extrapolated the temperature change, ΔT, of the reacting solution from the graph. The value of ΔT was +4.75°C. I then calculated the heat absorbed by the solution by using the formula c x mass x ΔT:
Heat absorbed by solution = c x mass x ΔT = 4.18J/g x 50g x (+4.75) =
= +992.75J
I then calculated the mole fraction of NaHCO3 used: n= Mass / Mr :
Mr NaHCO3 = 84
Moles NaHCO3 = mass NaHCO3 = 3.203 = 0.038
Mr NaHCO3 84
Finally, I calculated the molar ΔH (the heat of solution per mole of solute) by using: heat absorbed / moles. This represents ΔH1 with reference to our reaction cycle:
Molar ΔH = heat absorbed by solution (J) = +992.75= +26125 J/mol-1 =
Moles NaHCO3 0.038
= +26.13 kJ/ mol-1
Procedure for Experiment 2: with sodium carbonate
Initially, I gathered all the necessary apparatus together and set up my calorimeter, as with the previous experiment. I ensured that the polystyrene cup was clean and intact, that the lid was secure and was pierced correctly so that I could fit a thermometer through the hole.
Using a 25 cm3 burette, I accurately measured 50cm3 of 2M hydrochloric acid, and carefully poured this into my calorimeter. I then put the thermometer into place and recorded the temperature of the acid at 30 s intervals for the next three minutes. I recorded the results of this in a table (please see separate sheet).
I then accurately weighed the plastic weighing bottle on scales accurate to three decimal places. The weight came to 6.136g. I then added 2.0g of dry anhydrous sodium carbonate, ensuring that the weight of Na2CO3 was slightly over the stated 2.0g. This will help to minimise error, as some of the Na2CO3 might be left in the weighing bottle when we add it to the acid, or could stick to the sides of the calorimeter, whereas we want to ensure that all of the stated 2.0g of Na2CO3 react with the acid. The weight of the bottle plus the Na2CO3 was 8.261g.
I then added the dry anhydrous Na2CO3 to the 50 cm3 of 2M HCl in the polystyrene cup, and quickly stirred it to ensure the reactants were completely homogenised. I then recorded the temperature of the solution at 30 s intervals as it rose steadily, until the temperature stabilised. This took about ten minutes. I added these results to the aforementioned table (please see separate sheet).
I weighed the empty weighing bottle once again, (having emptied the Na2CO3 into the calorimeter), and the weight was 6.136g. I then deducted the result of this from the second weight (bottle + 2.0g Na2CO3): this will give us a more accurate figure for the actual mass of Na2CO3 used, as it accounts for possible error in any material left in the weighing bottle. I found that the actual mass of Na2CO3 I used was 2.125g.
After having collected all the results in my table, I used them to plot a graph of temperature against time for this reaction between Na2CO3 and HCl. I then extrapolated the temperature change, ΔT, of the reacting solution from the graph. The value of ΔT was -4.75°C. I then calculated the heat absorbed by the solution by using the formula c x mass x ΔT:
Heat absorbed by solution = c x mass x ΔT = 4.18J/g x 50g x (-4.75) =
= -992.75J
I then calculated the mole fraction of Na2CO3 used: n= Mass / Mr :
Mr Na2CO3 = 106
Moles Na2CO3 = mass Na2CO3 = 2.125 = 0.020
Mr Na2CO3 106
Finally, I calculated the molar ΔH (the heat of solution per mole of solute) by using: heat absorbed / moles. This represents ΔH2 with reference to our reaction cycle:
Molar ΔH = heat absorbed by solution (J) = -992.75= - 49637.5 J/mol-1 =
Moles Na2CO3 0.020
= -49.64 kJ/ mol-1
Calculating ΔH3:
Now that we have the values for ΔH1 and ΔH2 we can calculate the value of ΔH3 from these.
Looking at our cycle we can say: ΔH1 = ΔH3 + ΔH2 but we need to find the value of ΔH3. Therefore: ΔH3 = ΔH1 - ΔH2
Using our calculated values of molar enthalpy for ΔH1 and ΔH2, we can write our calculation of ΔH3 as follows:
ΔH3 = ΔH1 - ΔH2 = +26.13 – (- 49.64) = +75.77kJ/mol-1
Therefore the reaction for ΔH3 is highly endothermic.
Conclusion & Evaluation
Precision of measurements
Experimental error is always with us; it is in the nature of scientific measurement that uncertainty is associated with every quantitative result. This may be due to inherent limitations in the measuring equipment, or of the measuring techniques, or perhaps the experience and skill of the experimenter. It goes without saying that, when conducting this experiment, I endeavoured to be as careful and as accurate as I could in my measurements. However, there are several areas where precision could be improved:
-
When measuring out the 2M HCl, I had to use a 25 cm3 burette instead of a 50 cm3 one. This means that I had to measure two 25 cm3 quantities of acid into the calorimeter, which in turn could double the possibility of error. Enthalpy is an extensive property the magnitude of ΔH depends on the quantity of reactant, so measuring the correct quantity of acid is important.
-
The thermometer I used was accurate only to 0.5°C, so it was not always possible to gauge precisely any small changes in temperature.
-
I tried to minimise errors when measuring the dry ingredients (NaHCO3 and Na2CO3). I attempted to ensure that the weight of NaHCO3 and that Na2CO3 was slightly over the stated 3.0g and 2.0g recommended for them respectively. This will help to minimise error, as some of these reactants might be left in the weighing bottle when we add them to the acid, or could stick to the sides of the calorimeter, whereas we want to ensure that all of product reacts with the acid. Once again, the magnitude of ΔH depends on the quantity of reactant, so measuring the correct quantities of NaHCO3 and Na2CO3 is important.
- The calorimeter was not perfectly insulated and heat could enter or leave the system from the surroundings. Heat could be dispersed or gained from the atmosphere in the laboratory, and the thermometer also absorbs some heat. The calorimeter could have been better insulated if two cups nested together were to have been used, sealed by a tight-fitting cork lid, as follows:
-
Calibrations for instruments are made under certain conditions, which have to be reproduced if the calibrations are to be true within the specified limits. Volumetric apparatus is usually calibrated for 20oC, for example; the laboratory is usually at some other temperature.
-
I have tried to restrict all calculations to a sensible and meaningful number of significant figures, as a consequence of the possible errors and inaccuracies mentioned above. In addition, given the basic equipment used in this experiment, it would be meaningless for me to calculate a figure for ΔH to the nearest thousandth kJ/mol-1, for example.
Extrapolation of ΔT
We have seen how the insulation of the calorimeter system is not perfect and there is some heat loss or gain to the surroundings.
If the system is at room temperature initially, the initial temperature of the system will be stable. The final temperature, however, typically will not be at room temperature. If the reaction is exothermic, the final temperature will be above room temperature, and heat will slowly be lost from the calorimeter to the surroundings. Instead of being constant, the final temperature will slowly drift downward. The converse is true for an endothermic reaction.
To compensate for this heat loss/gain, the temperature is measured as a function of time, (please see graphs). Prior to the initiation of the chemical reaction, the temperature versus time plot can be used to establish the initial baseline. When the reaction is complete, the temperature versus time plot can be used to establish the final baseline. These baselines are typically linear, especially if the calorimeter is well insulated, which makes the heat loss/gain relatively small. The baselines represent the temperature changes attributable to heat exchange from the surroundings.
To determine Ti and Tf, (initial and final temperatures), we can imagine that the reaction took place instantaneously and the heat from the reaction is instantly distributed evenly throughout the system. In this case, the temperature would instantaneously jump from Ti to Tf at the time the reaction was initiated. These temperatures cannot be measured directly, but they can be inferred by extrapolating the baselines to obtain the temperatures on the initial and final baselines at the time the reaction was initiated. In this way, we obtain ΔT for both reactions. We must remember, however, that the graphs are dependent on accurate temperature readings, and I have already mentioned that the thermometer used was accurate only to 0.5°C.
Signs of ΔT and ΔH
Chemical reactions involve breaking some bonds and forming new ones. Breaking bonds absorbs energy (it is endothermic), whereas bond forming liberates it (it is exothermic). If the total energy needed to break bonds exceeds that liberated when the new ones form, the reaction will be endothermic. If, however, the total energy needed to break bonds is smaller than that liberated when the new ones form, the reaction will be exothermic.
We can also say that if the reaction products contain less energy than the reactants, the surplus energy is given to the surroundings in the form of heat. This means that the reaction is exothermic and ΔH is negative, (as in experiment 2). ΔH is negative because the initial temperature of the reactants is lower than the final temperature, and ΔT is therefore also negative.
If the reaction products contain more energy than the reactants, the surplus energy is absorbed by the system and the temperature of the reaction solution tends to drop. This means that the reaction is endothermic and ΔH is positive (as in experiment 1). ΔH is positive because the initial temperature of the reactants is higher than the final temperature, and ΔT is therefore also positive.
Estimating values of ΔT and ΔH
Before completing my experiments, I roughly estimated the value of ΔT for experiment 1 (NaHCO3 and HCl) as being approximately +4°C, and therefore estimated ΔH to be about + 22 kJ/mol-1. I consequently predicted that this reaction would be endothermic.
I then estimated the value of ΔT for experiment 2 (Na2CO3 and HCl) as being approximately - 4°C, and therefore estimated ΔH to be about - 41.8 kJ/mol-1. I consequently predicted that this reaction would be exothermic.
Using the reaction cycle and applying Hess’s Law, I predicted that the enthalpy change for the reaction
2 NaHCO3 Na2CO3 + H2O + CO2
would be endothermic and would measure about + 64 kJ/mol-1. My actual result was +75.77 kJ/mol-1 so my rough estimate was not too wide of the mark.
Calculations & Measurements - at a glance
Experiment 1: NaHCO3 and HCl
Weight empty bottle = 6.136g (W1)
Weight bottle plus 3.0g NaHCO3 = 9.339g (W2)
Weight empty bottle after NaHCO3 = 6.136g (W3)
Actual mass of NaHCO3 used = W2 – W3 = 3.203g
Mr NaHCO3 = 84
Moles NaHCO3 = mass NaHCO3 = 3.203 = 0.038
Mr NaHCO3 84
ΔT = +4.75°C (extrapolated from graph – please see separate sheet)
Heat absorbed by solution = shc x mass x ΔT = 4.18J/g x 50g x 4.75°C =
= +992.75 J
Molar ΔH = heat absorbed by solution = +992.75 = +26125 J/mol-1 =
Moles NaHCO3 0.038
= +26.125 kJ/mol-1
Experiment 2: Na2CO3 and HCl
Weight empty bottle = 6.136g (W1)
Weight bottle plus 2.0g Na2CO3 = 8.261g (W2)
Weight empty bottle after Na2CO3 = 6.136g (W3)
Actual mass of Na2CO3 used = W2 – W3 = 2.125g
Mr Na2CO3 = 106
Moles Na2CO3 = mass Na2CO3 = 2.125 = 0.020
Mr Na2CO3 106
ΔT = -4.75°C (extrapolated from graph – please see separate sheet)
Heat absorbed by solution = shc x mass x ΔT = 4.18J/g x 50g x -4.75°C =
= -992.75J
Molar ΔH = heat absorbed by solution = -992.75 = -49637.5 J/mol-1 =
Moles Na2CO3 0.020
= -49.638 kJ/ mol-1
