CHM 101: LABORATORY
Instructor Name: Dr Aisha Al Ashi
Student Names: Mariana Mneimne, Miriam Kazoun, Ayah Omairi, and Laila Fattouh
Student IDs: 202203343, 202203661, 202306149, and 202307004
Date of Performance: 31 October 2023.
Chemistry is an essential study of life. Through chemistry, the capability of
understanding various reactions requires accurate observations. Energy is either absorbed or
released during a chemical reaction/
A chemical reaction is the process through which a product’s property and composition
differs from that of a reactant.
Enthalpy (H) is the total energy within a system.
●
ΔH is the enthalpy change in a chemical reaction (from reactants to products).
○ ΔH= H of products - H of reactants.
○ If ΔH is less than zero, products H < reactants H; the reaction is called
exothermic.
■ Exothermic: releases heat, producing energy as heat.
○ If ΔH is more than zero, products H > reactants H; the reaction is called
endothermic.
■ Endothermic: absorbs energy, reactants need to be supplied with
energy to form products.
Enthalpy change in acid/base reactions is an exothermic reaction where heat is of
neutralization.
Hess’s Law: enthalpy change remains the same whether a reaction is done directly or
undergoes numerous steps.
-
Ex: reaction 1= reaction 2 + reaction 3, then, ΔH1 = ΔH2 + ΔH3/
Note: when we reverse reactions, the ΔH needs to be reversed (- becomes +, vice versa).
Procedure & Materials:
Materials: Reaction 1
●
●
●
●
●
●
Beaker.
50 mL of NaOH (1M, concentration).
45 mL Distilled Water.
Thermometer
5 mL of H2SO4 (10M, concentration).
Stirring rod.
Procedure: Reaction 1
1. Put 50 mL of NaOH into a beaker.
2. Measure 45 mL of distilled water into the Beaker. Measure its temperature (T1) using a
thermometer.
3. Measure H2SO4 and take its temperature (T2).
4. Add the measured H2SO4 to the beaker of NaOH and distilled water.
5. Stir the mixture for 30 seconds and record its temperature (T3).
Material: Reaction 2
●
●
●
●
●
●
90 mL distilled water.
Beaker.
Thermometer.
10 mL H2SO4 (10M, concentration).
Stirring rod.
Procedure: Reaction 2: Dilution of H2SO4
1.
2.
3.
4.
Place the distilled water into a beaker and record its temperature (T4).
Measure the H2SO4, and record its temperature (T5).
Add the measured H2SO4 into the beaker.
Stir the mixture for 30 seconds and record its temperature.
Material: Reaction 3
●
●
●
50 mL of NaOH (1M, concentration).
50 mL of H2SO4 (1M, concentration).
Stirring Rod.
Procedure: Reaction 3:
1. Put the NaOH and record its temperature (T7).
2. Record the temperature (T8) of H2SO4, then place it into the beaker.
3. Stir for 30 seconds and record the temperature (T9).
Analysis: Experiment 1
H2SO4 (T2) + NaOH (T1) → NaHSO4 (T3)+ H2O (exothermic)
●
●
●
Water + NaOH: 25 degrees celsius (T1)
H2SO4: 37 degrees celsius (T2).
After stirring the whole mixture: 39 degrees celsius (T3)
H2SO4 → NaHSO4: T3 -T2 → needs Q (H2SO4)
●
Q (H2SO4) = m(C)(ΔT)
○ m = d/v
■ m = (1g/mL, given) / (5mL) = 0.2 g.
○ C = 4.18 J/g.C
○ Q= (0.2g) x (4.18 J/g.C) x (39 - 37) = 1.672 J.
NaOH → NaHSO4: T3 -T1 → needs Q (NaOH)
●
Q (NaOH) = m(C)(ΔT)
○ m = d/v
■ m = (1g/mL, given) / (50 mL) = 0.02 g.
○ C = 4.18 J/g.C
○ Q = (0.02 g) x (4.18 J/g.C) x (39 - 25) = 1.1704 J.
Q(total) = Q(H2SO4 ) + Q(NaOH)
= 1.672 + 1.1704 = 2.8424 J.
Number of moles = 0.05 mol in both H2SO4 and NaOH.
From the balanced equation, we can see that 2 moles of NaOH react with 1 mole of H2SO4 to
produce 2 moles of water. Therefore, the enthalpy change for the reaction is:
ΔH total = (ΔH(NaOH) + ΔH(H2O)) - (ΔH(H2SO4) + ΔH(NaHSO4)).
At STP,
The standard enthalpy of formation:
●
●
●
●
NaOH = - 470.11 kJ/mol.
H2O = - 285.83 kJ/mol.
H2SO4 = - 814.2 kJ/mol.
NaHSO4 = - 943.7 kJ/mol.
ΔHtotal = [(0 kJ/mol + 2 ( - 285.83 kJ/mol)) - ( - 814.2 kJ/mol + 0 kJ/mol)] * (0.05 mol)
ΔHtotal = ( - 571.66 kJ/mol + 814.2 kJ/mol) x (0.05 mol)
= 12.126 kJ
Analysis: Experiment 2
H2SO4 (T5) + H2O (T4) → H2SO4 (T6) (dilution)
●
●
●
Water: 25 degrees celsius
H2SO4: 32 degrees celsius
After stirring entire mixture: 19 degrees celsius
●
Q(H2O) = m(C)(ΔT)
○ M = d/v
■ m= (1g/ml, given) / 90mL = 0.01 g
○ C = 4.18 J/g.C
○ Q = (0.01 g) x (4.18 J/g.C) x (19 - 25) = - 0.2508 J
●
Q(H2SO4) = m(C)(ΔT)
○ M = d/v
■ m= (1g/ml, given) / 10mL = 0.1 g
○ C = 4.18 J/g.C
○ Q = (0.1 g) x (4.18 J/g.C) x (19 - 32) = - 5.434 J
Q1 = - (Q (H2O) + Q (H2SO4)) = - ( - 0.2508 + ( - 5.434)) = - 5.6848J
n= c(v) = 0.1 mol
ΔH1 = Q1 / n1 = - 5.6848 / 0.1 = - 56.848 J
Analysis: Experiment 3
H2SO4 (T8) + NaOH (T7) → NaHSO4 (T9) + H2O
●
●
●
NaOH: 28 degrees celsius
H2SO4: 26 degrees celsius
After stirring entire mixture: 60 degrees celsius
●
Q(H2SO4) = m(C)(ΔT)
○ M = d/v
■ m= (1g/ml, given) / 50mL = 0.02 g
○ C = 4.18 J/g.C
○ Q = (0.02 g) x (4.18 J/g.C) x (60 - 26) = 2.8424 J
●
Q(NaOH) = m(C)(ΔT)
○ M = d/v
■ m= (1g/ml, given) / 50mL = 0.02 g
○ C = 4.18 J/g.C
○ Q = (0.02 g) x (4.18 J/g.C) x (60 - 28) = 2.6752 J
Q2 = - (Q (H2SO4) + Q (NaOH)) = - (2.8424 + 2.6752) = - 5.5176J
n=0.05 mol
ΔH2 = Q2 / n2 = - 5.5176J / 0.05 = - 110.352 J
ΔH (total) = ΔH1 (from reaction 2) + ΔH2 (from reaction 3)
ΔH (total) = - 56.848 J + ( - 110.352 J) = 167.2 J
To sum up the whole experiment, we can see how in chemistry, some substances can go through
various reactions using or releasing different quantities of energy.
Errors may have occurred throughout any of these reactions due to several factors such as
systematic errors in which the true value is imprecise due to false experimental setup or
measurement techniques. Random errors can occur unexpectedly due to environmental or human
conditions/errors.