September 01, 2015
By the end of today's lesson you should
·know what enthalpy is
·
·
·understand the difference between a molar enthalpy value and an
enthalpy change
·
·
·
·be able to use a formula to calculate an enthalpy change
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Understanding Enthalpy Changes in Chemical Reactions: Calculations and Examples, Study Guides, Projects, Research of Chemistry
A lesson on enthalpy changes in chemical reactions, explaining the concepts of molar enthalpy and enthalpy change, and providing examples and formulas for calculating enthalpy changes. It covers endothermic and exothermic reactions, and includes examples of the combustion of ethane and the formation of sulfur trioxide.
What you will learn
- What is enthalpy and how does it relate to chemical reactions?
- What is the molar enthalpy of reaction for water?
- What is the difference between molar enthalpy and enthalpy change?
- What is the molar enthalpy of reaction for ethane?
- How can you calculate the enthalpy change for a chemical reaction?
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By the end of today's lesson you should
know what enthalpy is
· understand the difference between a^ molar enthalpy^ value and an
enthalpy change
· be able to use a formula to calculate an enthalpy change
Thermal energy is released or absorbed by chemical
reactions
Endothermic Reactions
· more energy is required to break bonds than is released
by bond formation
Exothermic Reactions
· more energy is released in bond formation that is required
to break bonds
Each chemical reaction has its own enthalpy change
∆H depends on:
-the reaction (products and reactants)
-the coefficients of the balanced equation
Example
N
2(g)
- 3 H
2(g)
--> 2 NH
3(g)
∆H = -90 kJ
2 N
2(g)
- 6 H
2(g)
--> 4 NH
3(g)
∆H =
Converting from an enthalpy change to a molar
enthalpy value
Example 1
2 KCl
(s)
+ 2 H
SO
4(l)
-‐-‐-‐-‐> 2 HCl
(g)
+ K
SO
4(s)
∆H = + 41.0 kJ
What is the molar enthalpy change for HCl
(g)
Example 2
2 NaOH (s)
- 2 Al (s)
+ 2 H
O
(l)
-‐-‐> 2NaAlO 2(aq)
+ 3 H
2(g)
∆H = -‐850 kJ
What is the molar enthalpy of reaction for hydrogen?
Converting from a molar enthalpy to an enthalpy
change
Converting from a molar enthalpy to an enthalpy
change
∆H = n ∆rH
Example
2 ZnS (s)
- 3 O 2(g)
-‐-‐-‐-‐> 2 ZnO (s)
- 2 SO 2(g)
∆ r
H (ZnS) = -‐ 439.1 kJ/mol
What is the enthalpy change for the reaction?
Example 2
SO 2(g)
- 2 H 2
S (s)
-‐-‐> 3 S (l)
- 2 H 2
O (g)
∆ r
H (S) = -‐31.0 kJ/mol
What is the enthalpy change for this reaction?
∆H = n∆rH
∆H = 3 mol S x - 31. 0 kJ/mol
∆H = - 93. 0 kJ
Converting from one molar enthalpy value to another
given a balanced equation
Example 1
C
H
O
11(s)
+ 12 O
2(g)
-‐-‐-‐> 12 CO
2(g)
+ 11 H
O
(l)
r
H (C
H
O
) = -‐5640.3 kJ/mol
What is the enthalpy change for the reaction of 20.0 g
of sucrose (C 12
H
O
Example 2
SO
2(g)
+ 2 H
S
(s)
-‐-‐> 3 S
(l)
+ 2 H
O
(g) ∆rH(S) = -‐ 31.0 kJ/mol
What is the enthalpy change for the reaction of 100 g of S(s)?
Calculating an enthalpy change for a given mass of reactant or
product given a balanced equation with an enthalpy change
step 1: Find the molar enthalpy for the substance you are given the mass of
step 2: Find the ∆H = n ∆rH, where n=m/M
Example 1
2 NaOH (s)
- 2 Al (s)
+ 2 H
O
(l)
-‐-‐> 2NaAlO 2(aq)
+ 3 H
2(g) ∆H = -‐850 kJ
How much energy is released when 250 grams of
aluminum reacts?
Example 2
2 KCl (s)
+ 2 H
SO
4(l)
-‐-‐-‐-‐> 2 HCl (g)
+ K
SO
4(s)
∆H = + 41.0 kJ
How much energy is required to produce 40.0 grams of
K
SO
Step 1: ∆rH = ∆H/n
∆H = +41.0 kJ/ 1 mol K 2
SO 4
∆H = +41.0 kJ/mol
Step 2: ∆H = n ∆r H
= 40.0 g/174.27 g/mol x +41.0 kJ/mol
∆H = +9.41 kJ
-3.94 x 10 3 kJ
Determining moles of reactant or product given an enthalpy
change
Example 1
C
H
O
11(s)
+ 12 O
2(g)
-‐-‐-‐> 12 CO
2(g)
+ 11 H
O
(l)
r
H (C
H
O
) = -‐5640.3 kJ/mol
If 10 000 kJ of energy is released, what mass of sucrose reacted?
Example 2
2 KCl
(s)
+ 2 H
SO
4(l)
-‐-‐-‐-‐> 2 HCl
(g)
+ K
SO
4(s)
∆H = + 41.0 kJ
If 300 kJ of energy is absorbed, how much KCl
(s)
reacted?
Step 1: Find ∆rH for KCl
∆rH = ∆H/n = +41.0 kJ/ 2 mol KCl
∆rH = +20.5 kJ/mol
n = ∆H / ∆rH
n = + 300 kJ / + 20.5 kJ/mol
n= 14.6 mol
m = 14.6 mol x 74.55 g/mol = 1076 g
m = 1.09 kg