WESTERN LEYTE COLLEGE

SENIOR HIGH SCHOOL- SCIENCE, TECHNOLOGY,

ENGINEERING AND MATHEMATICS (STEM)
K to 12 CURRICULUM

FIRST LAW
OF
THERMODYNAMICS

Instructor: Elton Jhon M. Meridor

Principal: Isabelita T. Peroso

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General Chemistry I
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FIRST LAW OF THERMODYNAMICS

Objective
Explain the First Law of Thermodynamics

Overview
In this lesson, learners are expected to describe and explain the energy changes in chemical
reactions.

Background
Thermodynamics is the branch of Science which deals with the interconversion of various forms
of energy into one another. The underlying principle is the conservation of energy, the experimental
observation that energy cannot be created or destroyed. This principle can be used to assess the energy
changes that accompany physical and chemical processes.
System and Surroundings
The specified portion of the universe observation is called the system. A system is called
homogeneous if physical properties and chemical composition are identical throughout the system. A
heterogeneous system consists of parts each of which has different physical and possibly different
chemical properties also. The rest of the universe which is not a part of the system is called
surroundings.
Types of Systems
Three commonly encountered systems are defined as:
1. Isolated system – a system which can exchange neither energy nor matter with its surroundings.
2. Closed system – a system which can exchange energy but not matter with its surroundings.
3. Open system – a system which can exchange matter as well as energy with the surroundings.
Types of Processes
A process means a change in at least one of the state variables (measurable properties that
depend only on the state of the system). It gives a path or operation by which a system changes from one
state to another. Some common types of processes are:
1. Isothermal process – a process in which the temperature of the system remains constant.
2. Adiabatic process – a process in which no heat can leave or enter the system.
3. Isobaric process – a process in which the pressure of the system remains constant.

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4. Cyclic process – a process which consists of several steps, but the system returns to its original
state after undergoing various changes.
5. Reversible process – a process in which the direction may be reversed at any stage by a small
change in a state variable. The driving force is only infinitesimally greater than the opposing
force and is carried out infinitesimally slowly.
6. Irreversible process – a process which is not reversible. The magnitude of the driving force is
very different from the opposing force. All natural processes are irreversible.
Work, Heat, and Energy
Work, heat, and energy have the same units. Energy is a thermodynamic property of the system.
The total energy of the system is equal to the internal energy, U, the energy due to the position and
energy due to the motion of the system as a whole. Work and heat are not properties of the system but
meaningful when a system undergoes a process. The units of energy are erg, calorie, or joule.
The various common units of energy are related as follows:
1 cal = 4.184 x 102 erg = 4.184 J
Work
Work is equal to the force multiplied by the distance through which it acts. Mechanical work is
performed when a system changes its volume against an opposing pressure. If a gas expands against a
constant external pressure, pex, and its volume changes by an infinitesimally small amount, dV, then the
small amount of work done, dw, on the system is:
w = - pexdV
Heat
The heat absorbed by the system is expressed as q. It is another form of energy which the system
can exchange with the surrounding and flows from higher to lower temperatures. When a system
undergoes a change of state, the values of q and w for the process depend on the path.
Sign Conventions for Work and Heat
Processes Sign
Work done by the system on the surroundings -
Work done on the system by the surroundings +
Heat absorbed by the system from the surroundings (endothermic process) +
Heat absorbed by the surroundings from the system (exothermic process) -

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Heat Capacity and Specific Heat

The energy absorbed and released as heat in a chemical or physical change is measured in a

calorimeter. Heat can be thought of as the energy transferred between samples of matter because of a

difference in their temperatures. Energy transferred moves from a higher temperature to a lower

temperature. The quantity of energy transferred as heat during temperature change depends on the nature

of the material changing temperature, the mass of the material changing temperature, and the size of the

temperature change. Specific heat, cp, is the amount of energy required to raise the temperature of one

gram of substance by one Celsius degree or one Kelvin. It is mathematically expressed as:

q
cp =
mx ∆T

Examples:
1. Calculate the work required to raise a mass of 500 g through a height of 1 m (g = 9.8 m s-2)
w = mgh
w = 500 g = 0.5 kg g = 9.8 m s-2 h=1m
w = (0.5 kg) x (g = 9.8 m s-2) x (1 m)
w = 4.9 kg m2 s-2
w = 4.9 J

One mole of a gas occupying 3 dm3 is expanded against a constant external pressure of 1 atm to a
volume of 15 dm3. Calculate the work done on the system.
w = - pexdV
w = - pex (V2 - V1)
w = - (1 atm) x (15 dm3 - 3 dm3)
w = - 12 atm dm3 (1 atm dm3 = 101.32 J)
w = - 12 x 101.32 J
w = - 1215.84 J
w = - 1.216 KJ

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(The negative value indicates that the work is done by the system)
2. A 4.0 g sample of glass was heated from 274 K to 314 K, a temperature increase of 40 K, and
was found to have absorbed 32 J or energy as heat.
a. What is the specific heat of this type of glass?
b. How much energy will the same glass sample gain when it is heated from 314 K to 344 K?
Solution:
q
a. cp =
mx ∆T

32 J
cp =
( 4.0 g) x (314 K−274 K)

cp = 0.20 J/gK

b. q = cp x m x ∆ T
q = (0.20 J/gK) x (4.0 g) x (344 K – 314 K)
q = 24 J
First Law of Thermodynamics
It states that energy can neither be created nor destroyed although it can be transformed from one
form into another. The law can also be stated as:
(i) The energy of an isolated system remains constant.
(ii) It is not possible to construct a perpetual-motion machine which can work endlessly
without the expenditure of energy.
It can be mathematically expressed as:
Infinitesimal Change ∆U = ∆q + ∆w (q is heat gained)
Finite Change ∆U = q + w (w is the work done)
Energy is a state function and ∆U depends on initial and final states, whereas q and w depend on
the path. However, whatever may be the process, ∆U is always equal to q + w if the initial and final
states are well defined.
Example:
Two hundred and eighty grams of nitrogen absorbs 100 cal of heat without change in volume.
Calculate the ∆U.
Solution:
∆U = q + w

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q = 100 cal w = 0 (no change in volume)

∆U = 100 cal + 0
∆U = 100 cal
Enthalpy of a System
The heat of reaction is the amount of energy released or absorbed as heat during a chemical
reaction. Enthalpy, H, is traditionally known as the heat measurement of a system. However, the content
of enthalpy cannot be directly measured, it is only the changes in enthalpy that can be measured. An
enthalpy change (∆H) is the amount of energy absorbed or lost by a system as heat during a process at
constant pressure. The enthalpy change is always the difference between the enthalpies of the products
and the reactants. It expressed as:
∆H = H products – H reactants
Enthalpy is also defined as:
H = U + pV
The change in enthalpy is also defined as:
∆H = ∆U + p∆V at constant pressure process
∆H = ∆U + p∆V
= q -p∆V + p∆V
∆H = qp (enthalpy change is equal to heat absorbed)

Hess’s Law
The overall enthalpy change in a reaction is equal to the sum of enthalpy changes for individual
steps in the process. According to this principle, the enthalpies of reactions can be added or subtracted
algebraically.
Example:
Calculate the heat of formation of methane gas.
C (s) + 2H2 (g) CH4 (g) ∆Hf = ______

Solution:
C (s) + O2 (g) CO2 (g) ∆H = -393.5 kJ/mol
1
H2 (g) + O2 (g) H2O (l) ∆H = -285.8 kJ/mol
2
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H = -890.8 kJ/mol

Principles to be considered in combining thermochemical equations.

1. If the reaction is reversed, the sign of ∆H is also reversed.

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2. Multiply the coefficients of the known equations so that when added together will give the
desired thermochemical equation.

Thermochemical equation 3 is reversed:

CO2 (g) + 2H2O (l) CH4 (g) + 2O2 (g) ∆H = +890.8 kJ/mol
Thermochemical equation 2 is multiplied by a factor of 2:
1
2 [H2 (g) + O2 (g) H2O (l) ∆H = -285.8 kJ/mol]
2
2H2 (g) + O2 (g) 2H2O (l) ∆H = -571.6 kJ/mol

Add the three equations:

C (s) + O2 (g) CO2 (g) ∆H = -393.5 kJ/mol
2H2 (g) + O2 (g) 2H2O (l) ∆H = -571.6 kJ/mol
CO2 (g) + 2H2O (l) CH4 (g) + 2O2 (g) ∆H = +890.8 kJ/mol
______________________________________________________________________
C (s) + 2H2 (g) CH4 (g) ∆Hf = -74.3 kJ/mol

REFERENCES

CALIBO, C., A. ACABAL, J. G. JANSALIN, A. ACABAL, O. CORALES, E. QUEVEDO, M. P.

LORETO, A. RAMAL, F. SALAS, E. LANDERITO, and D. A. VARRON. 2010. Laboratory
manual in Chemistry 11: General Chemistry. Revised Ed. Visayas State University, Visca,
Baybay City.

CHANG, R. 2010. Chemistry, 10th ed. McGraw-Hill Companies, Inc.

SMITH, J. 2011. Organic Chemistry, 3rd ed.

USC Chemistry Review Center. 2016. Chemistry Review Manual.

WHITTEN, K., DAVIS, R., PECK, M. L. and G. STANLEY. 2010. Chemistry, 9th ed.

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Name: ______________________________ Section: ________________________________

Date Received: _______________________ Date Submitted: _________________________
Score: ____________
Instructions: Answer the following problems by showing a step-wise solution and enclosed in a box the
final answer.
1. Two moles of an ideal gas at 2 atm and 27 ˚C is compressed isothermally to one half its volume
b an external pressure of 4 atm. Calculate the work done.

2. Work done during the expansion of a gas from 4 dm 3 to 6 dm3 against a constant external
pressure of 2 atm has been used to heat 1 mol of water. What is the final temperature of water if
initially it was 293 K? The specific heat of water at 293 K is 4.184 J/gK.

3. A given sample of nitrogen gas is weighing 2.8 g at 27 ˚C and 20 atm pressure. It was allowed to
expand isothermally against a constant external pressure of one atmosphere. Calculate ∆U, q, and
w. Assume ideal behavior of the gas.

4. Calculate the heat that must be supplied to raise the temperature of 2 kg of water from 25 ˚C to
its boiling point at atmospheric pressure. The average specific heat of water in the range 25 ˚C -
100 ˚C is 1.0 cal/gK. How long will a 2 kW heater take to supply this energy? (1 kW = 1000 J/s)

5. Determine the specific heat of a material if a 35 g sample absorbed 48 J as it was heated from
298 K to 313 K.

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6. Calculate the heat of reaction for the combustion of nitrogen monoxide gas, NO, to form
nitrogen dioxide gas, NO2, as given in the following thermochemical equations:
1
NO (g) + O2 (g) NO2 (g) ∆H = ?
2
1 1
N2 (g) + O2 (g) NO (g) ∆H = +90.29 kJ/mol
2 2
1
N2 (g) + O2 (g) NO2 (g) ∆H = +33.2 kJ/mol
2

7. Calculate the heat of formation of pentane, C5H12, using the information on heats of formation.
Solve by combining the known thermochemical equations.
C (s) + O2 (g) CO2 (g) ∆H = -393.5 kJ/mol
1
H2 (g) + O2 (g) H2O (l) ∆H = -285.8 kJ/mol
2
1 5
C5H12 (g) + 4O2 (g) CO2 (g) + 3H2O (l) ∆H = -1767.8 kJ/mol
2 2

8. Using the balance chemical equation and the following information below, calculate the heat of
formation for sulfur dioxide, SO2, from its elements, sulfur and oxygen.
1 3 1
S (s) + O2 (g) SO3 (g) ∆H = -197.6 kJ/mol
2 4 2
2SO2 (g) + O2 (g) 2SO3 (g) ∆H = -198.2 kJ/mol

9. When 2.61 grams of dimethyl ether, CH 3OCH3, is burned at constant pressure, 82.5 kJ of heat is
given off. Find ∆H for the reaction.

10. When aluminum metal is exposed to atmospheric oxygen (as in aluminum doors and windows),
it is oxidized to form aluminum oxide. How much heat is released by the complete oxidation of
24.2 grams of aluminum at 25 °C and 1 atm? The thermochemical equation is:
4Al (s) + 3O2 (g) 2Al2O3 (s) ∆H = -3352 kJ/mol reaction

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General Chemistry I