1.4 Energetics
Introduction to Energetics
a.
b.
c.
f.
demonstrate an understanding of the term enthalpy change, ΔH
construct simple enthalpy level diagrams showing the enthalpy change
recall the sign of ΔH for exothermic and endothermic reactions, eg illustrated by
the use of exo- and endothermic reactions in hot and cold packs
evaluate the results obtained from experiments using the expression:
q = mcΔT
and comment on sources of error and assumptions made in the experiments. The
following types of experiments should be performed:
i. experiments in which substances are mixed in an
insulated container and the temperature rise measured
Connector –
Draw energy level diagrams for an exothermic and an endothermic
reaction. On your diagram show the activation energy and the
energy change for the reaction.
Crowe 2010
Exothermic & Endothermic reactions
Exothermic reaction
Energy is released.
Reaction vessel becomes hotter.
Temperature inside reaction
vessel increases.
Endothermic reaction
Energy is absorbed.
Reaction vessel becomes cooler.
Temperature inside reaction
vessel decreases.
Enthalpy
• The heat content of a chemical system is called the enthalpy (H)
• The enthalpy change (ΔH) is the amount of heat released or
absorbed when a chemical reaction occurs at constant pressure.
ΔH = H (products) – H (reactants)
units kJ mol-1 (kJ/mol)
• If ΔH is negative the reaction is exothermic
• If ΔH is positive the reaction is endothermic.
• Energy changes are measured under standard laboratory conditions
(Room Temperature Pressure - RTP):
25oC (298K) & 101.3kPa (1 atmosphere)
Practical 1.4 – Measuring some enthalpy changes
See separate slide show
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.
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
Enthalpy Change in Chemical Equations
• If a chemical equation is reversed, the sign of ΔH is also reversed.
• The value of ΔH given as kJ mol-1 refers to
kJ per 1 mole of reactant or product as written in the equation.
N2(g) + 3H2(g) -----> 2NH3(g)
ΔH = - 92.4 kJ/mol
The formation of ammonia is exothermic reaction
(ΔH is negative) and
92.4 kJ of energy per mole of nitrogen gas is released.
2NH3(g) -----> N2(g) + 3H2(g)
ΔH = + 92.4 kJ/mol
The decomposition of ammonia is an endothermic reaction
(ΔH is positive), and
92.4kJ of energy per mole of nitrogen gas is absorbed by the
reaction.
Manipulating the Enthalpy Change Term
The value of ΔH given as kJ mol-1 refers to kJ per 1 mole of reactant
or product as written in the equation.
N2(g) + 3H2(g) -----> 2NH3(g)
H = - 92.4 kJ/mol
• 92.4 kJ of energy is released for every 1 mole of N2(g)
• 92.4 kJ of energy is released for every 3 moles of H2(g)
• 92.4 kJ of energy is released for every 2 moles of NH3(g) produced.
1. How much energy is released if only 1 mole of ammonia (NH3) gas
is produced?
2. How much energy is released if 10 moles of nitrogen (N2) gas and
30 moles of hydrogen (H2) gas is used in the reaction?
3. How much energy is released if 5 moles of hydrogen (H2) gas and
5/3 mole of nitrogen (N2) gas is used in the reaction?
46.2 kJ, 924kJ, 154 kJ
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