Mass of the container after the crystals were added = 11.63g
Mass of the crystals that did not react = 00.15g
Mass of the crystals that reacted = 03.85g
UNCERTAINITIES
+/- 0.01g : Digital weighing scale
+/- 0.01s : Stop watch
+/- 0.05cm3 : Measuring cylinder
+/- 0.05°C : Thermometer
OBSERVATIONS
- When hydrochloric acid is added to sodium carbonate, some effervescence (bubbles appear) is observed because of the liberation of carbon dioxide gas. It is completely soluble and in the process it starts getting warmer.
- The same thing happens when sodium hydrogencarbonate is added to hydrochloric acid except for the fact it cools down instead of getting warmer.
CHEMICALS (QUALITATIVE DATA)
- Hydrochloric acid- It is colorless and odorless. It is a monoprotic acid, that is, it produces 1 hydrogen ion when completely dissolved in water. The molarity of hydrochloric acid used in this experiment is 2M.
- Sodium carbonate- Sodium Carbonate is a white, crystalline compound soluble in water (absorbing moisture from the air) but insoluble in alcohol. It forms a strongly alkaline water solution. It is also known as soda ash
-
Sodium hydrogencarbonate- sodium bicarbonate or sodium hydrogen carbonate, chemical compound, NaHCO3, a white crystalline or granular powder, commonly known as bicarbonate of soda or baking soda. It is soluble in water and very slightly soluble in alcohol.
DATA PROCESSING AND PRESENTATION
Finding the enthalpy of reaction for the following equation:
2NaHCO3 Na2CO3 + H2O +CO2
- Using 1.88g of sodium carbonate and 3.3g of sodium hydrogencarbonate
Enthalpy cycle for the reaction
∆H1
2NaHCO3(s) + 2 HCl 2NaCl(aq) +2 CO2(g)+ 2 H2O(l)
∆H2 ∆H3
Na2CO3(s) + CO2(g) + H2O + 2 HCL(aq)
Heat released during the reaction between HCl and NaHCO3 (Q1)= Mc∆T
So, Q1 = 25g * 4.18 * (15-23)
Q1 = 25* 4.18 * -8
Therefore, Q1 = -836 J
= -0.836KJ
Calculating the number of moles present in 3.3g of NaHCO3
Number of moles = mass(g)/ molar mass
Mass (g) = 3.3g
Molar mass = 23 + 1 + 12 +(16*3)
= 23 + 1 + 12 + 48
= 84 g/mol
number of moles present in 3.3g of NaHCO3 = 3.3g/ (84g/mol)
= 0.039 moles
Calculating the amount of energy given out by 1 mole
If 0.039 moles of NaHCO3 give -0.836KJ of energy then
1 mole would give out (-0.836/0.039 = -21.44KJ) of energy
Therefore, ∆H1 = -21.44 KJ/mol
Calculating ∆H3 by the above method, that is, the reaction between Na2CO3 and HCl
Q2 = Mc∆T
So, Q2 = 25 * 4.18 * (30-23)
Q2 = 25 * 4.18 * 7
Therefore, Q2 = 731.5J
= 0.7315KJ
Calculating the number of moles present in 1.8g of Na2CO3
Number of moles = mass (g)/ molar mass
Mass (g) = 1.8g
Molar mass = (23 * 2) + 12 +(16*3)
= 46 +12 + 48
= 106 g/mol
number of moles present in 1.8g of Na2CO3 = 1.8g/ (106g/mol)
= 0.018 moles
Calculating the amount of energy given out by 1 mole
If 0.018 moles of Na2CO3 give -0.731KJ of energy then
1 mole would give out (-0.7315/0.018 = 40.64KJ) of energy
Therefore, ∆H3 = 40.64 KJ/mol
In order to find the enthalpy of reaction for:
2NaHCO3 Na2CO3 + H2O+ CO2; we use the Hess’s law which states that 2 ∆H1 = ∆H2 + ∆H3
∆H2 = 2 ∆H1 - ∆H3
so, 2 ∆H1 = 2 * -21.44 KJ/mol
= -42.88KJ/mol
∆H3 = 40.64 KJ/mol
Therefore, ∆H2 = -42.88 – 40.64
= -83.52KJ/mol
-
Calculating the enthalpy change of reaction using 5.66g of sodium hydrogencarbonate, 3.85g of sodium carbonate and 50cm³ of HCl.
Enthalpy cycle for the reaction
∆H1
2NaHCO3(s) + 2 HCl 2NaCl(aq) +2 CO2(g)+ 2 H2O(l)
∆H2 ∆H3
Na2CO3(s) + CO2(g) + H2O + 2 HCL(aq)
Heat released during the reaction between HCl and NaHCO3 (Q1)= Mc∆T
So, Q1 = 50g * 4.18 * (21.5-23.0)
Q1 = 50* 4.18 * - 1.5
Therefore, Q1 = -331.50 J
= -0.3135KJ `
Calculating the number of moles present in 5.66g of NaHCO3
Number of moles = mass (g)/ molar mass
Mass (g) = 5.66g
Molar mass = 23 + 1 + 12 + (16*3)
= 23 + 1 + 12 + 48
= 84 g/mol
number of moles present in 5.66g of NaHCO3 = 5.66g/ (84g/mol)
= 0.0674 moles
Calculating the amount of energy given out by 1 mole
If 0.0674 moles of Na2CO3 give -0.3135KJ of energy then
1 mole would give out (-0.3135/0.0674 = -4.65KJ) of energy
Therefore, ∆H1 = -4.65 KJ/mol
Calculating ∆H3 by the above method, that is, the reaction between Na2CO3 and HCl
Q2 = Mc∆T
So, Q2 = 50 * 4.18 * (28.5-23)
Q2 = 50 * 4.18 * 5.5
Therefore, Q2 = 1149.5J
= 1.1495KJ
Calculating the number of moles present in 3.85g of Na2CO3
Number of moles = mass (g)/ molar mass
Mass (g) = 3.85g
Molar mass = (23 * 2) + 12 + (16*3)
= 46 +12 + 48
= 106 g/mol
number of moles present in 1.8g of Na2CO3 = 3.85g/ (106g/mol)
= 0.036 moles
Calculating the amount of energy given out by 1 mole
If 0.036 moles of Na2CO3 give 1.1495KJ of energy then
1 mole would give out (1.1495KJ/0.036 = 31.93KJ) of energy
Therefore, ∆H3 = 31.93KJ/mol
In order to find the enthalpy of reaction for:
2NaHCO3 Na2CO3 + H2O+ CO2; we use the Hess’s law which states that 2 ∆H1 = ∆H2 + ∆H3
∆H2 = 2 ∆H1 - ∆H3
so, 2 ∆H1 = 2 * -4.65 KJ/mol
= -9.3KJ/mol
∆H3 = 31.93KJ/mol
Therefore, ∆H2 = -9.3KJ/mol - 31.93KJ/mol
= -41.23KJ/mol
- Error analysis
Digital weighing scale: 1) 0.01/3.3 * 100
= ±0.3%
2) 0.01/5.66 * 100
= ±0.18%
Stop watch : 0.01/300 * 100
= ± 3.3*10^-3
Measuring cylinder : 1) 0.05/25 * 100
= ±0.2%
2) 0.05/50 * 100
= ±0.1%
Thermometer : 0.05/23 * 100
= ±0.22%
Total percentage error = 0.22%+ 0.1% + 3.3*10^-3 + 0.18%+ 0.3%
= ±0.8033%
Accounting for the error ∆H2 = ± -41.23KJ/mol
= ± -83.52KJ/mol
Conclusion
The reaction between sodium hydrogencarbonate and HCl is endothermic, that is, heat is being absorbed in the reaction and the reaction between sodium carbonate and HCl is exothermic because temperature is given out to the surroundings.
Also, in the second part of the experiment when the volume of HCl is increased and also the masses of sodium hydrogencarbonate and
sodium carbonate is increased, the temperature difference in the reaction is less than before when the mass were less. The enthalpy of reaction is also decreased in the second part.
Evaluation:
Reasons for shortcoming in the answers are as follows:
- While transferring the HCl from the measuring cylinder to beaker, there is a possibility of leaving out some amount of HCl in the beaker itself.
- An analogue thermometer was used so the temperature may not have been accurate.
- Some systematic errors in the equipment might have led to some slight changes in the readings.
Solution to the above problems:
- The first problem stated above is a personal error; So it can only be overcome by practice and improving one’s concentration while doing the experiment.
- The second problem could have been overcome by the use of digital thermometer which is more accurate than an analogue thermometer.