Chemical Energetics &
Thermodynamics Q5
Jeremy Koo
1
Question Summary
Aim of experiment: to
determine ΔHneuto of
reaction
Aqueous HCl +
Aqueous NaOH
Volume: 50cm3
2
Concentration:1
mol/dm3
Stand for 1 hour
to reach lab
temperature
Temperature
taken every 30s
for 2.5min
At 3min,
solutions were
mixed
Temperature
taken again
Part a(i)
 Temperature correction
Maximum temperature assuming
no heat loss
= 22.0oC
Maximum temperature
rise
= 22.0oC - 15.0oC
= 7.0oC
3
Plot a best fit line using the
points after 3 minutes and
extrapolate it to intersect the line
Time = 2.5 min
Extrapolate the temperature
recorded at Time = 2.5min
upwards
Part a(ii)
 Total volume of solution = 2 x 50.0 cm3
= 100 cm3
 Assuming all solutions to have a density of 1.0 g/cm3
 Total mass of solution = 100 cm3 x 1.0 g/cm3
= 100 g
 Specific heat capacity = 4.2 J g-1 K-1
 Change in temperature = 7.0 oC
4
Part a(ii)
Energy
transferred
Specific heat capacity = 4.2
J g-1 K-1
 q = m c ΔT
Mass of solution
= 100g
Temperature
change = 7.0 oC
 q = 100g x 4.2 J g-1 K-1 x 7.0oC
5
= 2940 J
= 2.94 kJ
Part a(ii)

ΔHneuto
=
−𝑞
𝑚
=
−2.94 𝑘𝐽
1 𝑥 0.0500 𝑚𝑜𝑙
= -58.8 kJ/mol
6
Part b(i)
 Original setup (HCl)
0.0500 mol
0.0500
HClmol
(0.0500
NaCl
mol H+)
7
0.0500 mol
0.0500
mol
NaOH
H2O
(0.0500
mol OH-)
Part b(i)
 New setup (H2SO4)
0.0500 mol
0.0500
H2SOmol
4
(0.100
NaClmol
H+)
8
0.0500 mol
H2O
0.0250
0.0500 mol
H
NaOH
2SO4
(0.0500
molOH
H+)-)
mol
Part b(i)
Temperature rise would still be 7.0oC
• 1 mol H2SO4 dissociates to form 2 mol H+
• Only half the volume of acid is required to neutralize NaOH
However, the total number of moles of H+ and OH- that react are
still the same
ΔHneuto remains the same
Temperature rise remains the same
9
Part b(ii)
Original experiment
• HCl (aq) + NaOH (aq)
• ΔHneuto = -58.8 kJ/mol
NaCl (aq) + H2O (l)
New experiment
• HCl (aq) + NaOH (s)
• ΔH = -99.8kJ/mol
NaCl (aq) + H2O (l)
• NaOH (s)
Na+ (aq) + OH- (aq)
• ΔHsolno = -44.4 kJ/mol
• Energy released to dissolve NaOH pellets
• Reaction more exothermic
10
Thank you
Comments?
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