Chemistry Exam 3

Front 1

The energy of position or composition.
In chemistry it is
refers to the energy stored in bonds.

Back 1

Potential Energy

Front 2

The energy of motion. In chemistry, it is referred
to as molecular motion.

Back 2

Kinetic Energy

Front 3

The sum of all the energy in a system.

Back 3

Internal Energy

Front 4

The higher the
temperature,

Back 4

the higher
the kinetic energy.

Front 5

the amount of
thermal energy (kinetic
energy) transferred between
two objects at different
temperatures.

Back 5

Heat

Front 6

the potential energy that is stored in the bonds
that make up a molecule.

Back 6

Chemical Energy

Front 7

Molecules with high energy bonds, when converted to low energy
bonds will release chemical energy.

Back 7

This energy is given off as
heat or as work.

Front 8

A common reaction used to
release energy is the
combustion of butane gas.

Back 8

C4H10 (g) + O2 (g)

Front 9

Energy is not produced
or destroyed in ordinary chemical reactions. Energy can
only be converted from one form into another. Also
known as the Law of Conservation of Energy.

Back 9

First Law of Thermodynamics

Front 10

ΔEuniverse = 0 = ΔEsystem + ΔEsurroundings

Back 10

System – What is being studied
• Surroundings – Everything else in the Universe
• Universe – System + Surroundings

Front 11

Energy flow from the system to the surroundings
has a

Back 11

negative sign (loss of energy).

Front 12

Energy flow from the surroundings to the system
has a

Back 12

positive sign (gain of energy).

Front 13

ΔEuniverse = 0 = ΔEsystem + ΔEsurroundings

Δ Esystem = - ΔEsurroundings

Back 13

know this

Front 14

If the surroundings get hotter then heat must have

Back 14

emanated from the
system.

Front 15

If the surrounding get cold then heat must have

Back 15

been absorbed from the
surroundings

Front 16

Δ Esystem = - ΔEsurroundings

Back 16

Δ Esystem = heat + work = q + w

Front 17

If the temperature increases energy
as heat must have originated in the
system.

Back 17

The system lost energy as
heat so q is -.

Front 18

If the temperature decreases
energy as heat must have been
absorbed by the system.

Back 18

The
system gained energy as heat so q
is +

Front 19

If ΔV is positive then –P ΔV is
negative.

Back 19

Work is negative and
system lost energy as work.

Front 20

If the ΔV is negative so -P
ΔV is positive

Back 20

Work is
positive and system gained
energy as work.

Front 21

If ΔV = 0 then w = 0
If ΔP = 0 then w = -PΔV
1 liter atm = 101 Joules

Back 21

know this

Front 22

In order to calculate work at constant pressure the
following equation is applied.
w= -P Δ V
work will only be appreciable when there are large
changes in volume of the system.

Back 22

If system grows substantially work is negative
if system shrinks substantially work is positive
if system remains essentially the same size – work is
negligible.

Front 23

The explosion of 2.00 mol of solid TNT with a volume of approximately
0.274 L produces gases with a volume of 489 L at room temperature.
How much PV (in kilojoules) work is done during the explosion?
Assume P = 1 atm, T = 25°C.
2 C7H5N3O6(s) → 12 CO(g) + 5 H2(g) + 3 N2(g) + 2 C(s)

Back 23

answer in note book last page plug in later.

Front 24

the science of measuring heat
changes (q) for chemical reactions.

Back 24

Calorimetry

Front 25

think of heat capacity as

Back 25

capacity to store heat.

Front 26

the amount of heat required to
raise the temperature of an object or substance a
given amount.

Back 26

Heat capacity (C)
the equation is
C =q / Δ T

Front 27

The amount of heat required to
raise the temperature of 1.00 g of substance by 1.00°C.

Back 27

Specific Heat Capacity:

Front 28

The amount of heat required to
raise the temperature of 1.00 mole of substance by
1.00°C.

Back 28

Molar Heat Capacity:

Front 29

What is the specific heat of lead if it takes 96 J to
raise the temperature of a 75 g block by 10.0°C?

Back 29

q = specific heat capacity(mass)delta T

96 J = specific heat
capacity(75g)(10.0 C)

96J / (75g)(10.0 C)= 0.128kJ/g C