1.4 Energetics
Practical 1.4 – Measuring some enthalpy changes
f. evaluate the results obtained from experiments using the expression:
energy transferred in joules = mass x specific heat capacity x temperature change
and comment on sources of error and assumptions made in the experiments.
Connector – explain the terms:
• redox reaction
• neutralisation reaction
• exothermic reaction
• endothermic reaction
Crowe2009
Useful information
 Usually one reagent is in excess to ensure a complete reaction
• So calculations should be based on the fully reacted reagent.
 Certain assumptions are made during the calculation
• The density of the solution and its specific heat capacity is
that of water.
• That no heat is lost to the surroundings.
 The main source of error in these experiments is
• heat loss to the surroundings – (atmosphere & equipment)
 Other sources of error include:
• incorrect measurements
• solution concentrations
• mass of reactants
Accuracy
• Remember to give numerical answers in line with that
of the least accurate measurement recorded – don’t
just write your calculator’s result with lots of decimal
places!
• % error = uncertainty in measurement x 100
reading
E.g.1 a balance has an uncertainty of 0.01 g when read to the
second decimal place. What is the % error for a reading of 2.64g?
% error = 0.01 x 100 = 0.38% Note: measurements to
2dp so answer to 2dp
2.64
E.g.2 the temperature rise was 6.1oC, while the thermometer could
be read to ± 0.5oC. What was the % error?
(Hint: how many readings are required to determine a temperature change?)
% error = 2 x 0.5 x 100 = 16.4%
6.1
Note: measurements to
1dp so answer to 1dp
Error in measuring temperature changes can be reduced by using
graphs produced manually, or by a data logger:
A data series is plotted and
the max temp change is
obtained by extrapolation.
The data is plotted in real
time and the max temp
change is obtained directly.
Working out enthalpy changes in solutions
Part 1 Calculating the heat absorbed by the solution
q= m
c
x
x
ΔT
E.g. When 50cm3 of 1.0M HNO3 was neutralised by 50cm3 of 1.0M NaOH
the following results were obtained:
Starting temperature of solns.
= 19.0oC
Maximum temperature reached = 25.1oC
Calculate the heat absorbed by the solution, given that:
the mass of 1cm3 of water = 1g;
and that the specific heat capacity of water = 4.2 JK-1g-1.
q= m
x
c
x
ΔT
q = 100
x
4.2
x
6.1 = 2562 J
Working out enthalpy changes in solutions
Part 2 Calculating the enthalpy change
1. Write a balanced equation for the reaction between nitric
acid and sodium hydroxide.
2. What are the units for ΔH?
3. What is the definition of the enthalpy of neutralisation?
HNO3 (aq) + NaOH (aq)
NaNO3 (aq) + H2O (l)
kJ mol -1
Enthalpy of neutralisation is the enthalpy change
when 1 mole of H+ ions from an acid is neutralised by
1 mole of OH- ions
Use the information given in the question and your
answer from q = mcΔT to work out the enthalpy of
neutralisation for the above reaction.
Solution
50cm3 of 1.0M HNO3 was neutralised by 50cm3 of 1.0M NaOH
Energy = 2562 J = 2.562 KJ
Enthalpy of neutralisation is the enthalpy change when 1 mole of
H+ ions from an acid is neutralised by 1 mole of OH- ions
No. of moles of H+ ions used = 1.0 x 50 = 0.05 moles
1000
moles
Concentration x volume
in dm3
Solution
50cm3 of 1.0M HNO3 was neutralised by 50cm3 of 1.0M NaOH
Energy = 2562 J = 2.562 KJ
Enthalpy of neutralisation is the enthalpy change when 1 mole of
H+ ions from an acid is neutralised by 1 mole of OH- ions
No. of moles of H+ ions used = 1.0 x 50 = 0.05 moles
1000
So 0.05 moles of H+ ions produced 2.562 KJ energy
1.0 mole of H+ ions will produce 1.0 x 2.562 KJ energy
0.05
ΔH neut = - 51.2 KJ mol-1
Why is the ΔH value –ve?
Why is the answer to 1dp?
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