3. Is a branch of chemistry that describes
the energy changes that occur during a
chemical reaction. It is part of
thermodynamics, which studies the
relationship between chemical reactions
and energy.
4. is a thermodynamic quantity equivalent
to the total heat content of a system.
5. is a reaction that absorbs energy in the form
of heat. As a result, endothermic reactions
feel cool to the touch and can even lower the
temperature of the surroundings.
6. is a reaction that releases energy in the form
of heat. As a result, exothermic reactions
feel hot to the touch and can even raise the
temperature of the surroundings.
7. Enthalpy is a thermodynamic quantity that represents the total
heat content of a system. In a chemical reaction, the enthalpy
change (∆H) refers to the difference in enthalpy between the
products and reactants. Enthalpy change is the amount of heat
absorbed or released by a system during a chemical reaction.
The heat absorbed or released during a reaction is dependent on
the type of reaction. There are two types of reactions:
endothermic and exothermic. An ENDOthermic reaction is a
reaction that absorbs heat, while an EXOthermic reaction is a
reaction that releases heat.
8. Endothermic Reaction: Photosynthesis
During photosynthesis, plants absorb energy from the sun to
convert carbon dioxide and water into glucose and oxygen. This
reaction is endothermic since energy is absorbed.
Exothermic Reaction: Combustion
During combustion, a fuel reacts with oxygen to produce carbon
dioxide and water while releasing heat and light. This reaction is
exothermic since energy is released.
9. Enthalpy of a reaction (∆Hrxn) is an extensive property that
depends on the amount of reactants consumed in the reaction.
The enthalpies of formation (∆H°f) are used to determine the
enthalpy of reaction at standard conditions, 1 atm and 25°C.
The equation for calculating the Enthalpy of a reaction is:
∆Hrxn = ∑Hproducts – ∑Hreactants
10. The equation for calculating the Enthalpy of a reaction is:
∆Hrxn = ∑Hproducts – ∑Hreactants
If the product has a higher enthalpy than the reactant, then the
value of ∆H is positive, indicating that the reaction is endothermic.
In an endothermic reaction, the heat content of the products is
greater than the heat content of the reactants because heat is
absorbed.
If the product has a lower enthalpy than the reactant, then the
value of ∆H is negative, indicating that the reaction is exothermic.
In an exothermic reaction, the heat content of reactants is greater
than the heat content of the products since heat is released.
11. What is the enthalpy change for the combustion of 1 mole of
methane (CH4) at constant pressure, given the following equation:
CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l); ∆H° = -180 kJ/mol
Given:
CO2(g): H° = -393.509 kJ/mol
H2O(l): H° = -241.818 kJ/mol
CH4(g): H° = -74.52 kJ/mol
O2(g): H° = 0 kJ/mol
17. Final answer: ∆Hrxn= -801.625 kJ/mol
Based on the calculated value of ∆Hrxn, which is -890.36 kJ/mol, we can
conclude that the combustion of 1 mole of methane is an exothermic reaction.
This is because the value of ∆Hrxn is negative, indicating that energy is released
from the reaction to the surroundings.
19. states that the total enthalpy change for
a chemical reaction is the same, no
matter how the reaction takes place.
20. the sum of the kinetic and potential
energies of the particles in a system.
21. the transfer of thermal energy between
two systems that are at different
temperatures.
22. Hess's law is named after Germain Hess, a Russian Chemist and Doctor.
According to Hess's law, the total enthalpy change for a reaction, regardless of
multiple stages, is the sum of all changes.
Enthalpy is a state function, allowing us to calculate the overall change by
summing up changes for each step until the product is formed.
Two principles must be understood to use Hess's law effectively:
Reversing an equation reverses the sign of the ΔH value.
Multiplying an equation by a coefficient multiplies the ΔH value by the same
coefficient.
23. We have three chemical reactions here. The last reaction have a missing enthalpy,
how can we determine its enthalpy change using the first two reactions?
2H2O → 2H2 + O2 ΔH= +572 KJ/mol
O2 → 2O ΔH= +495 KJ/mol
H2 + O → H2O ΔH= ?
We can do that by using Hess's Law!
24. HOW TO SOLVE IT:
The H20 on the first reaction is opposite to the H20 on the third reaction.
The 0 on the second reaction is opposite on the third reaction.
Divide the first and second reaction by 2.
Reverse the first and second reaction and also its ΔH.
Add those two reactions.
2H2O → 2H2 + O2 ΔH= +572 KJ/mol
O2 → 2O ΔH= +495 KJ/mol
H2 + O → H2O ΔH= ?
25. First, we reverse and divide the reactions. When reversing the reaction the sign of the ΔH also
reverses.
(2H2O → 2H2 + O2 ΔH= +572 KJ/mol) divided by 2
(O2 → 2O ΔH= +495 KJ/mol) divided by 2
Then, we cancel the similar terms
(1/2)O2 + H2 → H2O ΔH= -286 KJ/mol
O → (1/2)O2 ΔH= -247.5 KJ/mol
And when we combine the two reactions we will get the third reaction.
H2+O → H2O
Now we can get the change in enthalpy by adding the enthalpy on the first and second reaction
ΔH= -286 KJ/mol + ΔH= -247.5 KJ/mol = -533 KJ/mol
Therefore, using Hess’s Law we determined the change in enthalpy of the third reaction
H2+O → H2O ΔH= -533 KJ/mol
27. is the area of chemistry concerned with
the speed, or rates at which a chemical
reaction occurs.
28. is the speed at which a chemical reaction
takes place, defined as proportional to the
increase in the concentration of a product
per unit time and to the decrease in the
concentration of a reactant per unit time.
Reaction rates can vary dramatically.
29. a substance that enters into and is altered
in the course of a chemical reaction.
30. A rate law shows how the rate of a
chemical reaction depends on reactant
concentration.
31. is the minimum energy required to cause
a process (such as a chemical reaction) to
occur.
32. 1. NATURE OF THE REACTANTS - The rate of reactions depends on the nature of
the participating substances or the particular reactants and the number of bonds t
that have to be broken or formed.
Example:
33. 2. CONCENTRATION - concentration is an important
factor because molecules must collide in order to react
with each other. When the concentration of the
reactants increases, the frequency of the molecules
colliding increases, striking each other more frequently.
Rate law:
rate = k (A)¹ - (A)↑ rate↑
34. 3. SURFACE AREA - The surface area of a solid reactant has an
important effect on the rate of reaction.The rate of reaction will be
limited by the surface area of the phases that are in contact.
4. CATALYST - are substances used to facilitate reactions but remain
chemically unchanged afterwards. The rate of reaction is increased
when the catalyst provides a different reaction mechanism to occur
with lower activation energy.
Ea↓ K↑ rate↑
35. 5. TEMPERATURE - Molecules at a higher
temperature have more thermal energy
and collision frequency is greater at
higher temperatures.
37. is a mathematical expression that show
the rate of reactant depending on the
concentration of the reactant. It is
expressed as:
Rate = k [A]^×
38. Rate = k [A]ˣ
• k is the rate constant
• The power x is the order of the reaction with respect to
the reactant A
• A is the reactant
• The concentration of reactant or product is represented in
square bracket [ ]
• The half-life ( t½ ) of a reaction is the time it takes for an
initial amount of a reactant to be reduced to half
39. A zero- order reaction (x=0) has a constant rate that is equal to its
rate constant regardless of changes in the concentration of the
reactant, it is expressed as
Rate = k [ A ]⁰ or Rate = k
The half-life for zero-order reaction is given by
t½ = [ A ]o/2k
The concentration of the reactant at any given time of the reaction
can be obtained using the equation.
[ A ] = - kt + [ A ]o
40. Example:
The initial concentration of a reactant in a zero-order reaction is 0.75 M.
The rate constant k is 0.018 M/min
a) what will be the concentration of the reaction after 17 minutes?
b)How long will it takes the concentration to be reduced to 0.06 M?
a) [A]բ =- kt + [A]o
[A]բ = -0.018 M /min ( 17 min ) + 0.75 M
[A]բ = 0.44 M
41. Example:
The initial concentration of a reactant in a zero-order reaction is 0.75 M.
The rate constant k is 0.018 M/min
a) what will be the concentration of the reaction after 17 minutes?
b)How long will it takes the concentration to be reduced to 0.06 M?
b). [A] = -kt + [A]o
0.06 M = -0.018 M/min (t) + 0.75 M
0.06 M - 0.75 M = -0.018M/min (t)
-0.69 M = - 0.018 M/min(t)
-0.69M /-0.018M/min = 0.018M/min/0.018M/min(t)
t = 38.33 min.
42. In a first-order reaction ( x = 1 ), the rate doubles when the
concentration of a reactant is double. In general the rate increase
in the same order as the concentration of the reactant.
Rate = k [ A ] ¹ = k [ A ]
The half-life for first-order reaction is given by
t½ = 0.693/k
The integrated rate law is
In[A] = -kt + In[A]o
43. Dinitrogen pentoxide, N₂O₅ decomposes via first-order kinetics
with a rate constant of 0.16 s⁻¹
(a) what is the half-life in seconds for this reaction.
(b) If the initial concentration of N₂O₅ is 0.65 M , what
will be the concentration of it after 2 seconds.
(a). t½ = 0.693/k
t½ = 0.693/ .16
t½ = 4.33 s
44. Dinitrogen pentoxide, N₂O₅ decomposes via first-order kinetics
with a rate constant of 0.16 s⁻¹
(a) what is the half-life in seconds for this reaction.
(b) If the initial concentration of N₂O₅ is 0.65 M , what
will be the concentration of it after 2 seconds.
(b) In[A] = -kt + In[A]o
ln[A] = - 0.16/s (2s) + 0.65 M
ln[A]= -0.75 M
e^ln[A] = e^ (-0.75 )
[A] = 0.47 M
45. For a second-order reaction ( x = 2 ), increasing the reactant
concentration to twice as much quadruples the original reaction
rate. A triple increase in concentration means a ninefold increase
in reaction rate.
Rate = k [ A ]²
The half-life for second-order reaction can be derived to form
t½ = 1/ k[ A ]o
The integrated rate law equation
1/[A]=kt + 1/[A]o
46. Example:
The isomerization of ammonium cyanate ( NH4CNO ) to urea follows the second-order
kinetics
with a rate constant of 9.9x10⁻⁴/M•s at 25⁰ C. From an initial concentration of 0.24,
a. how much of it in M remains after 15 minutes?
b. What is the half-life in minutes for the reaction?.
Note: Convert the 15 min. to second and that equivalent to 900s
a). 1/[A] = kt + 1/[A]o.
1/ [A] = (0.0009 /M•s)(900s) + 1/0.24 M
1/[A] = 0.891 M + 1/0.24 M
= 5.1 /M
[A] = 0.20 M
47. Example:
The isomerization of ammonium cyanate ( NH4CNO ) to urea follows the second-order
kinetics
with a rate constant of 9.9x10^-4/M•s at 25⁰ C. From an initial concentration of 0.24,
a. how much of it in M remains after 15 minutes?
b. What is the half-life in minutes for the reaction?.
Note: Convert the 15 min. to second and that equivalent to 900s
b) t½ = 1/k[A]o
= 1/0.0002376
t½ = 4209 s or 70 min.
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