- Created by: chunks-42
- Created on: 17-04-15 16:13
Chemical Reactions Usually Have Enthalpy Changes
When chemical reactions happen, there'll be a change in energy. The souped-up chemistry term for this is enthalp change:
Enthalpy change is the heat energy transferred in a reaction at constant pressure. The units of enthalpy change are kJmol-1.
You write enthalpy change to show that the elements were in their standard states and that the measurements were made under standard conditions. Standard conditions are 100 kPa pressure and a stated temperature. In this section, all the enthalpy changes are measured at 298 K (25 degress Celsius).
Reactions can be either Exothermic or Endothermic
Exothermic reactions give out energy. Enthalpy change is negative.
In exothermic reactions, the temperature often goes up.
Oxidation is exothermic, e.g. the combustion of a fuel like methane or the oxidation of carbohydrates, such as glucose in respiration.
Endothermic reactions absorb energy. Enthalpy change is positive.
In these reactions, the temperature often falls. The thermal decomposition of calcium carbonate is endothermic. The main reactions of photosynthesis are also endothermic - sunlight supplies the energy.
Reactions are all about Breaking and Making Bonds
When reactions happen, reactant bonds are broken and product bonds are formed.
1. You need energy to break bonds, so bond breaking is endothermic (Enthalpy change is positive). Stronger bonds take more energy to break.
2. Energy is released when bonds are formed, so this is exothermic (enthalpy change is negative). Stronger bonds release more energy when they form.
3. The enthalpy change for a reaction is the overall effect of these two changes. If you need more energy to break bonds than is released when bonds are made, enthalpy change is positive. If it's less, enthalpy change is negative.
Mean Bond Enthalpies are not Exact
Water has got two O-H bonds. You'd think it'd take the same amount of energy to break them both... but it doesn't.
The first bong, H-OH: +492 kJmol-1
The second bond, H-O: +428 kJmol-1
So the mean bond enthalpu is 492+428/2= +460kJmol-1.
Breaking bonds is always an endothermic process, so mean bond enthalpies are always positive.
The data book says the bond enthalpy for O-H is 463 kJmol-1. It's a bit different because it's the average for a much bigger range of molecules, not just water. For example. it includes the O-H bonds in alcohols and carboxylic acids too.
Enthalpy Changes Can be Calulated using Average Bo
In any chemical reaction is absorbed to break bonds and given out during bond formation. The difference between the energy absorbed and released is the overall enthalpy change of reaction.
Enthalpy Change of Reaction= Total Energy Absorbed - Total Energy Released
Example: Calculate he overall enthalpy change for this reaction: N2 + 3H2-> 2NH3. Use the average bond enthalpy values: N---N= 945, H-H= 436, N-H=391.
Bonds broken: (1x945) + (3x436)= 2253 kJmol-1
Bonds formed: 6x391= 2346 kJmol-1
Enthalpy Change of Reaction= 2253-2346= -93 kJmol-1
* If you can't remember which value to subtract from which, just take the smaller number from the bigger one then add the sign at the end- positive if 'bonds broken' was the bigger number (endothermic), negative if 'bonds formed' was bigger (exothermic)
There are Different Types of Enthalpy Change
Standard enthalpy change of formation is the enthalpy change when 1 mole of a compound is formed from its elements under standard conditions.
E.g. 2C+ 3H2+1/2O2-> C2H5OH
Standard enthalpy change of combustion is the enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions.
Practice Questions Q1
Q1. The followng values are bond enthalpy values: C-H=435, C=O= 805, O=O=498, O-H= 464
The complete combustion of methane can be represented by the following equation:
a. Use the values of bond enthalpies above to calculate the enthalpy change for the reaction. (4marks)
b. Is the reaction endothermic or exothermic? Explain your answer. (1 mark)
Practice Question, Q2
Q2. Methanol, CH3OH (-762kJmol-1), when blended with petrol can be used as a fuel
a. Write an equation, including state symbols, for the standard enthalpy change of combustion of methanol. (2 marks).
b. Write an equation, including state symbols, for the standard enthalpy change of formation of methanol (2 marks)
c. Liquid petroleum gas is a fuel that contains propane, C3H8.
Explain why the following equation does not represent a standard enthalpy change of combustion. (1 mark)
2C3H8 + 10O2--> 8H20 +6CO2 (Enthalpy Change= -4113 kJmol-1)
Question 1 answers
1a. Bonds Broken= (4x435)+ (2x498) =2736 kJmol-1 (1 mark)
Bonds formed= (2x805) + (4x464) = 3466 kJmol-1 (1 mark)
Net energy change = 2736+ (- 3466) = -730 kJmol-1 (1 mark)
[ 1 mark for correct numerical value, 1 mark for correct units]
b. The reaction is exothermic, because the enthalpy change is negative/ more energy is given out than is taken in (1 mark)
Question 2 answers
2a. CH3OH (l)+ 1 1/5 O2 (g)--> CO2 (g)+ 2H2O (l)
Correct balanced equation (1 mark). Correct state symbols for reactants (1 mark)
It is perfectly OK to use halves to balance equations. Make sure that only 1 mole of CH3OH is combusted, as it says in the definition for the enthalpy change of combustion.
b. C (s) + 2H2 (g) -> CH3OH (l)
Correct balanced equation (1 mark). Correct state symbols for reactants (1 mark).
c. Only 1 mole of C3H8 should be shown according to the definition of the enthalpy change of combustion.
You really need to know the definitions of the standard enthalpy changes off by heart. There's loads of nit-picky little details they could ask you questions about.