Title: Chapter 15 - Standard enthalpy change of a reaction
1Chapter 15 - Standard enthalpy change of a
reaction
2Standard Enthalpy Changes Hana Amir and Madeley
3Definition
- Any reaction that depends on temperature,
pressure and state - The enthalpy change happens when all reactants
and products are in their standard state.
4Some enthalpies
- Enthalpy of reaction
- Enthalpy of formation
- Enthalpy of neutralization
- Enthalpy of hydration
- Enthalpy of combustion
- Enthalpy of solution
- Enthalpy of atomization
5Standard Enthalpy Changes of Reaction
- Definition
- The heat change when molar quantities of
reactants as specified by the chemical equation
react to form products at standard conditions - It depends in the physical state of reactants and
the products and the conditions under which the
reaction occurs. - Standard conditions are 298K (25o C) and
1.00105 Pa
6Standard Enthalpy change of Formation
- Enthalpy change that occurs when one mole of
substance is formed from its elements in the
standard state under standard conditions. - Standard conditions
- Temperature 298K (25o C)
- Pressure 1.00105 Pa
- 2C(graphite) 3H2(g) 1/2O2(g)-------gt C2H5OH
(I)
7- All elements in their standard states (oxygen
gas, solid carbon in the form of graphite, etc.)
have a standard enthalpy of formation of zero, as
there is no change involved. -
- Eg
- O(g) O(g) ----gtO2(g)
8Energy Cycle
9- From the diagram we get
- The chemical change elements to products
- can either occur directly or indirectly .
- The Total enthalpy change must be the same for
both routes. - S?H?f (Products) S?H?f (Reactants)
?H?reaction - This gives the general expression of
- ?H? reaction S?H?f (Products) - S?H?f
(Reactants)
10Example
- Calculate the enthalpy change for the reaction
- C3H8(g) 5O2(g)-------- 3CO2(g) 4H2O(g)
- Standard change of formation ?H?f /kJ mol-1
- C3H8(g) -105
- CO2(g) -394
- H2O(l) -286
11- Steps
- Write down the equation with the corresponding
enthalpies of formation underneath - C3H8(g) 5O2(g)-------- 3CO2(g) 4H2O(g)
- (-105) 0 3(-394)
4(-286) - Note The standard enthalpies of formation are
given in per mole , hence, they should be
multiplied by the numbers of moles in the balance
equation .
12- ?H?reaction S?H?f (Products) - S?H?f
(Reactants) - ?H?reaction 3(-394) 4(-286) -(-105)
- -2221 KJ mol-1
13Thermochemical Equations Balanced chemical
equation for a reaction including the enthalpy of
the reaction shown immediately after the
equation. In a thermochemical equation, the
coefficients represent moles and can therefore be
fractional. The following is an example IB Data
booklet gt -227 kj mol -1 Ethanol (C2H5OH) is
made from the elements (C (Graphite)) and
hydrogen (H2(g)) and oxygen (O2(g)) __C
(graphite) __H2 (g) ___O2(g)-----------gt
C2H5OH(I) ?H227kj mol -1 Balance the C, H and
0 2C (graphite) 3H2 (g) 1/2
O2(g)-------------- C2H5OH(I) ?H227 kj mol -1
14Questions!
1 .Use the table of standard enthalpies of
formation at 25C to calculate enthalpy change
for the reaction 4NH3(g) 5O2(g) ? 6H2O(g)
4NO(g)
2 .Write the thermodynamically equation for the
standard enthalpy of formation of propanone
enthalpy change CH3COCH3
15Answers! 1. 1031.76 kJ mol1 2. 3C (Graphite)
3H2(g) 1/2O2(g)-----------CH3COCH3
16Standard enthalpy change of Combustion
17What is standard enthalpy change of combustion?
- The standard enthalpy of combustion is the
enthalpy change that occurs when one mole of
substance burns completely under the standard
conditions of 25 ? and 1 atm. - Eg C6H14(l) 9O2(g) 6CO2(g)
7H2O(l) - The standard enthalpy of combustion is always
negative -
18Exercise!
- Write down the enthalpy of combustion equations
for the following reactions! - Methane
- CH4 (g) 2O2(g) CO2(g) 2H2O(l)
- Ethanol
- C2H5OH(l) 3O2(g) CO2(g)
2H2O(l) - Propane
- C3H8(g) 5O2(g) 3CO2(g)
4H2O(l)
19Calculating standard enthalpy change
Products
Reactants
Combustion Products
S?H?c(reactants) S?H?(products) ?H ?H
S?H?c(reactants) - S?H?c(products)
20Question!
May 2010 Paper 1 TZ 2B
21Question!
- Give an equation for the formation of glucose.
- 6C(graphite) 6H2 (g) 3O2(g)
C6H12O6(s) - Calculate the enthalpy of formation of glucose
- C ?H? -394 Kjmol-1 H2?H? -286
Kjmol-1 - C6H12O6 ?H? -2803 Kjmol-1
- ?H S?H?c(reactants) - ?H?c(products)
- ?H ( 6(-394) 6(-286) 3(0) ) - (-2803)
- -1277 Kjmol-1
22Question!
- Calculate the enthalpy change for the following
reaction! - C(s, graphite) C(s, diamond)
- C(s, graphite) O2(g) CO2(g) ?H?
-393 Kjmol-1 - C(s, diamond) O2(g) CO2(g) ?H?
-395 Kjmol-1 - Solution
?H?
C(s, graphite) O2(g)
C(s, diamond) O2(g)
-395 Kjmol-1
-393 Kjmol-1
CO2(g)
-393 Kjmol-1 -395 Kjmol-1 ?H? ?H 2
Kjmol-1
23Question
May 2008 Paper 1 TZ 1A
24Question
Nov 2007 Paper 1 D
25Comparison
Standard enthalpy of combustion
Standard enthalpy of formation
Products
Reactants
Combustion Products
?H S?H?c(reactants) - ?H?c(products)
Think First Price
Its not CPR in chem! Its CRP!
26Question!
May 2010 Paper 1 TZ 2A
27Questions!
May 2010 Paper 1 TZ1A
May 2010 Paper 1 TZ1C
28Question!
Nov 2009 Paper 1 TZ1C
Answers to all MCQ questions is the last letter
in the identification of the paper from which the
question was taken! )