ENERGY AND CHEMICAL REACTIONS - I
Name of teacher:
Week: Date: Time:
Period: Duration: 1 HR 20 MIN. Average age of learners: 16YEARS
Subject: CHEMISTRY Class: SS TWO
Topic: ENERGY AND CHEMICAL REACTIONS - I
Sub topic: INTRODUCTION TO ENERGY AND CHEMICAL
REACTIONS
Reference materials:
(1) ESSENTIAL CHEMISTRY, TONALD PUBLISHERS, I. O ODESINA
(2) NEW SCHOOL CHEMISTRY, AFRICAN FIRST PUBLISHERS, OSEI YAW ABABIO
(3) INTERNET
Instructional materials: Bomb calorimeter
Entry behavior: The students have been taught exothermic and endothermic reaction.
Behavioural objective: At the end of the lesson the students should be able to:
1. Explain heat of reaction
2. State the types of heat of reaction
3. State first and second laws of thermodynamic
4. Solve mathematical problems on enthalpy change, entropy and free energy
CONTENT
ENERGY AND CHEMICAL REACTIONS - I
Sub topic: INTRODUCTION TO ENERGY AND CHEMICAL
REACTIONS
Energy changes occur in chemical reactions as reactants change to products. This is because the reactants and the products of a given chemical reaction possess different amount of chemical energy present in a substance, but we can observe the energy changes that accompany chemical reactions in the form of:
· Heat- when a strong acid and a strong base are mixed.
· Light and heat- when magnesium is burned in air.
· Sound and heat- when a mixture of hydrogen and oxygen is ignited.
Of these energy changes, we can only measure the changes in heat energy with ease.
Heat content and Heat of reaction.
Every substance possesses a characteristic internal energy which is due to its structure and physical state. This is known as its heat content or enthalpy.
The standard heat of reaction is the amount of heat evolved or absorbed when a chemical reaction occurs between molar quantities of the substances as represented in the equation of reaction under standard conditions.
The basic unit for measuring heat change is calories or joule.
1 calorie = 4.18 joule (J).
From the Law of Conservation of Energy, the heat absorbed in decomposition a compound is equal to the heat evolved in its formation under the same conditions.
∆H (heat of reaction) = Heat contents of products – Heat of content of reactants.
∆H =
Hproducts - Hreactants
TYPES OF ENTHALPY CHANGE OR HEAT OF REACTION (∆H)
1. HEAT OF FORMATION: The energy change when 1 mole of a substance is formed from the element.
2. HEAT OF SOLUTION: The energy change when 1 mole of a substance is dissolved in large enough volume of solvent that dilution does not bring about further heat change.
3. HEAT OF COMBUSTION: The energy change when 1 mole of a substance is burned completely in oxygen under standard condition.
4. HEAT OF FUSSION: The energy change when 1 mole of substance is melted.
5. HEAT OF VAPOURIZATION: The energy change when 1 mole of a substance is vapourized.
6. HEAT OF NEUTRALIZATION: The energy change when 1 mole of H+ is neutralized by 1 mole of OH-.
CHEMICAL THERMODYNAMICS
Thermodynamics is the study of relationship between heat and other form of energy. Heat is represented by the symbol q. All forms of energy are referred to commonly as work in thermodynamics and are represented w. The relationship between heat and work is expressed in the laws of thermodynamics.
The first law of thermodynamics:
It state that energy may be converted from one form to another, but it cannot be created or destroyed. This is actually the Law of Conservation of Energy.
Change in internal energy (∆U) = Heat absorbed by the system (q) - work done by the system (w)
When the system is gaseous, work done is given by p∆V , p = pressure and ∆V = change in volume.
∆U = q + w
∆U = q + p∆V
The second law of thermodynamics:
It state that a spontaneous process occurs only if there is an increase in the entropy of a system and its surroundings i.e. the change in the total entropy of the system, ∆Stotal must be positive.
Enthalpy, entropy, and free energy are the three factors which play a role in determining whether a reaction occurs spontaneously.
Entropy ,S, is a measure of degree of disorder or randomness of a substance. The perfectly ordered system at 0 K has an entropy of zero. It is measure in J K-1 mol-1. The change in entropy of a system, ∆S, is given by ∆S =∆H/T, where ∆H is the enthalpy change, T is the absolute temperature.
If heat is absorbed, ∆S is positive and there is an increase in entropy. If heat is evolved, ∆S is negative and there is an decrease in entropy.
Free energy, G, of a system is that energy which is available for doing work. The change is the free energy of the system, ∆G, is given by ∆G = ∆H + T∆S. for all spontaneous, ∆G must be negative (∆G = - T∆Stotal).
For a chemical change to occur spontaneously:
I. ∆Stotal must be positive as the total entropy must increase.
II. ∆G must be negative.
To determine the molar heat of neutralisation for the
reaction:
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
Extract the data needed to calculate the molar heat of
neutralisation for this reaction:
V(NaOH) = volume of NaOH(aq) in the calorimeter
= 50.0 mL
V(H2SO4) = volume of H2SO4(aq) added
to achieve neutralisation = 25.0 mL
c(NaOH) = concentration of NaOH(aq) = 1.0 mol L-1
c(H2SO4) = concentration of H2SO4(aq) =
1.0 mol L-1
Ti = initial temperature of solutions before additions =
18.0°C
Tf = final temperature of solution at neutralisation =
26.9°C
d = density of solutions = 1 g mL-1 (assumed)
Cg = specific heat capacity of solutions = 4.18 J°C-1g-1 (assumed)
q = heat liberated during neutralisation reaction = ? J
Check the units for consistency and convert if
necessary:
Convert volume of solutions (mL) to mass (g):
density × volume = mass
since density = 1 g mL-1:
1 × volume (mL) = mass (g)
mass(NaOH) = 50.0 g
mass(H2SO4) = 25.0 g
Calculate the heat produced during the neutralisation
reaction:
heat produced = total mass × specific heat capacity × change in
temperature
q = mtotal × Cg × ΔT
mtotal = mass(NaOH) +
mass(H2SO4() = 50.0 + 25.0 = 75.0 g
Cg = 4.18 J°C-1g-1
ΔT = Tf - Ti = 26.9 - 18.0 = 8.9°C
q = 75.0 × 4.18 × 8.9 = 2790.2 J
Calculate the moles of water produced:
OH-(aq) + H+(aq) → H2O(l)
1 mol OH-(aq) + 1 mol H+(aq) → 1 mol H2O
moles(H2O) = moles(OH-(aq))
moles(OH-(aq)) = concentration (mol L-1)
× volume (L)
= 1.0 × 50.0/1000
= 0.050 mol
moles of water produced = 0.050 mol
Calculate the heat liberated per mole of water produced,
ΔHneut :
ΔHneut will be negative because the reaction is
exothermic
ΔHneut = heat liberated per mole of water
= -1 × q ÷ moles of water
ΔHneut = -1 × 2790.2 ÷ 0.050
= -55803 J mol-1 (of water produced)
We can divide J by 1000 to convert this enthalpy change to kJ per mole:
ΔHneut = 55803 J mol-1 ÷
1000 J/kJ
= 55.8 kJ mol-1 (of water produced)
Example 1
Calculate ΔH for the following reaction:
8 Al(s) + 3 Fe3O4(s) → 4 Al2O3(s) + 9 Fe(s)
Solution
ΔH for a reaction is equal to the sum of the heats of formation of the product compounds minus the sum of the heats of formation of the reactant compounds:
ΔH = Σ ΔHf products - Σ ΔHf reactants
Omitting terms for the elements, the equation becomes:
ΔH = 4 ΔHf Al2O3(s) - 3 ΔHf Fe3O4(s)
The values for ΔHf may be found in the Heats of Formation of Compounds table. Plugging in these numbers:
ΔH = 4(-1669.8 kJ) - 3(-1120.9 kJ)
ΔH = -3316.5 kJ
Example 2
Calculate ΔH for the ionization of hydrogen bromide:
HBr(g) → H+(aq) + Br-(aq)
Solution
ΔH for a reaction is equal to the sum of the heats of formation of the product compounds minus the sum of the heats of formation of the reactant compounds:
ΔH = Σ ΔHf products - Σ ΔHf reactants
Remember, the heat of formation of H+ is zero. The equation becomes:
ΔH = ΔHf Br-(aq) - ΔHf HBr(g)
The values for ΔHf may be found in the Heats of Formation of Compounds of Ions table. Plugging in these numbers:
ΔH = -120.9 kJ - (-36.2 kJ)
ΔH = -120.9 kJ + 36.2 kJ
ΔH = -84.7 kJ
Example 3
Calculate the standard heat of reaction for the reaction of nitrogen monoxide gas with oxygen to form nitrogen dioxide gas.
The balanced equation is:
solution
Applying the equation form the text:
The standard heat of reaction is -113 kJ.
The reaction is exothermic, which makes sense because it is a combustion reaction and combustion reactions always release heat.
Example 4
Determine if this reaction is spontaneous under standard conditions; knowing that the reaction's change is enthalpy is ΔH∘=−144 kJ, and its change in entropy is ΔS∘=−36.8 J/K.
4KClO3(s)→3KClO4(s)+KCl(s)
solution
We know that ΔG∘=ΔH∘−T⋅ΔS∘ for standard state conditions, which imply a pressure of 1 atm and a temperature of 298 K, so
ΔG∘=−144⋅103J−298K⋅(−38.6JK)=−133 kJ
If ΔG∘<0, the reaction is said to be spontaneous, so this reaction is spontaneous under standard conditions.
Let's determine for what temperatures this reaction will be spontaneous. In other words, we need to find a temperature at which the reaction stops being spontaneous.
A reaction is no longer spontaneous when ΔG>0, i.e. when ΔH−T⋅ΔS>0. So,
ΔH−T⋅ΔS>0→ΔH>T⋅ΔS→ΔHΔS>T
We get −144⋅103J−38.6JK=3732>T, which means that this reaction is spontaneous for any T<3731 K.
Example 5
Methane gas reacts with water vapor to produce a mixture of carbon monoxide and hydrogen according to the balanced equation CH4(g)+H2O(g)→CO(g)+3H2(g)
The ΔH∘ for the reaction is +206.1 kJ/mol, while the ΔS∘ is +215 J/K • mol. Calculate the ΔG∘ at 25°C and determine if the reaction is spontaneous at that temperature.
solution
ΔH∘=206.1 kJ/mol; ΔS∘=215 J/K⋅mol=0.215 kJ/K⋅mol; T=25∘C=298 K; ΔG∘=? kJ/mol
ΔG∘=ΔH∘−TΔS∘=206.1 kJ/mol−298 K(0.215 kJ/K⋅mol)=+142.0 kJ/mol
The resulting positive value of ΔG indicates that the reaction is not spontaneous at 25°C.
The unfavorable driving force of increasing enthalpy outweighed the favorable increase in entropy. The reaction will be spontaneous only at some elevated temperature.Available values for enthalpy and entropy changes are generally measured at the standard conditions of 25°C and 1 atm pressure. The values are slightly temperature dependent and so we must use caution when calculating specific ΔG values at temperatures other than 25°C. However, since the values for ΔH and ΔS do not change a great deal, the tabulated values can safely be used when making general predictions about the spontaneity of a reaction at various temperatures.
PRESENTATION
i.The teacher explains heat of reaction.
ii. The teacher states and explains types of heat of reaction .
iv. The students states types of heat of reaction.
iii. The teacher explains laws of thermodynamics.
iv. The teacher leads the students to solve problems on heat of reaction
EVALUATION
The teacher evaluates the lessons by asking the following questions:-
1. Explain heat of reaction
2. State the types of heat of reaction
3. State first and second laws of thermodynamic
4. Calculate ΔG⁰ for a reaction where ΔH⁰ is equal to 36.2 kJ and ΔS⁰ is equal to 123 J/K at 298 K. Is this a spontaneous reaction?
Solution
ΔG⁰ = ΔH⁰ – TΔS⁰
ΔG⁰ = 36.2 kJ – (298 K x 123 J/K)
ΔG⁰ = -0.4 kJ
Therefore the reaction is spontaneous because ΔG⁰ is negative.
ASSIGNMENT
Calculate the standard free energy change for the following reaction, using standard free energies of formation:
5 C(s) + 2 SO2(g) → CS2(g) + 4 CO(g)
Is this a spontaneous reaction?
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