735305
Unit 1 -
Enthalpy
- Enthalpy Changes
& Activation Energy
- every substance contains an
energy known as enthalpy (H)
- this enthalpy changes when
reactants turn into products
- the enthalpy change
(/\H) is the difference
in energy between
products & reactants
- the enthalpy change
(/\H) is the difference
in energy between
products & reactants
- this enthalpy changes when
reactants turn into products
- when reactant molecules
collide, they form a highly
energetic arrangement
called an activated complex
- the activation energy is the
minimum energy required
for reactant molecules to
form an activated complex
- the activated
complex can
form products
or turn back
to reactants
- Energy is
released when
new bonds are
formed in the
products
- Energy is
released when
new bonds are
formed in the
products
- the activated
complex can
form products
or turn back
to reactants
- the activation energy is the
minimum energy required
for reactant molecules to
form an activated complex
- every substance contains an
energy known as enthalpy (H)
- Potential Energy Diagrams
- exothermic
reactions
- heat is given out
- /\H =
Hproducts
-
Hreactants
- Hproducts
<
Hreactants
- therefore,
the enthalpy
change is
negative (-ve)
- therefore,
the enthalpy
change is
negative (-ve)
- Hproducts
<
Hreactants
- /\H =
Hproducts
-
Hreactants
- heat is given out
- endothermic
reactions
- heat is taken in
- /\H =
Hproducts -
Hreactants
- Hproducts
>
Hreactants
- therefore, the
enthalpy change
is positive (+ve)
- therefore, the
enthalpy change
is positive (+ve)
- Hproducts
>
Hreactants
- /\H =
Hproducts -
Hreactants
- heat is taken in
- exothermic
reactions
- Effect of a Catalyst
- catalysts
lower
the Ea
- more product can be formed as
more particles have the energy
needed to overcome the Ea barrier
- catalysts have
no effect on the
enthalpy change
- catalysts have
no effect on the
enthalpy change
- more product can be formed as
more particles have the energy
needed to overcome the Ea barrier
- catalysts
lower
the Ea
- Enthalpy of Combustion
- the
enthalpy of
combustion
is when 1
mole of a
substance
burns
completely
in oxygen
- energy
produced
by a
known
mass of
fuel
burning
can be
calculated
- Eh = cm/\T
- c = specific heat
capacity of water
- m = the mass
of water being
heated (kg)
- /\T = the temperature difference
in degrees in centigrade
- c = specific heat
capacity of water
- Eh = cm/\T
- the
enthalpy of
combustion
is when 1
mole of a
substance
burns
completely
in oxygen
- PPA 3
- determine the
combustion of ethanol
- burner
contains
ethanol
- burner
contains
ethanol
- enthalpy given for ethanol in data book is
higher than result due to heat loss in the
surroundings or incomplete combustion
- ethanol is flammable
- shouldn't be placed
near a Bunsen burner
- ethanol is flammable
- shouldn't be placed
near a Bunsen burner
- determine the
combustion of ethanol
- Enthalpy
of Solution
- the enthalpy of solution is the enthalpy change
when 1 mole of a substance dissolves in water
- Example: KOH(s) = K+(aq) + OH-(aq)
- Eh = cm/\T
= 4.18 x
0.1 x 12`C
= 5.01kJ
- M = n x GFM
= 1 x 56.1
= 56.1kJ
- 5g KOH = 5.01kJ
56.1g KOH
= 56.1 x .01/5
= 56.3kJ mol ^-1
- 5g KOH = 5.01kJ
56.1g KOH
= 56.1 x .01/5
= 56.3kJ mol ^-1
- M = n x GFM
= 1 x 56.1
= 56.1kJ
- Eh = cm/\T
= 4.18 x
0.1 x 12`C
= 5.01kJ
- Example: KOH(s) = K+(aq) + OH-(aq)
- the enthalpy of solution is the enthalpy change
when 1 mole of a substance dissolves in water
- Enthalpy of
Neutralisation
- the enthalpy of neutralisation is
the energy change when an acid is
neutralised to form 1 mole of water
- Example: HCl(aq) + NaOHkJ(aq) = NaCl(aq) = H2O(l)
- vol. H2O = vol. acid + vol. alkali = 100 + 100 = 200cm3 (0.2kg - 1cm3 H20 = 1g)
- Eh = cm/\T = 4.18 x 0.2 x 6`C = 5.02kJ
- mole ratio = 1:1
- moles of acid = c x V = 1 x 0.1 = 0.1moles => moles of H2O = 0.1moles
- 0.1moles H2O = 5.02kJ 1 mole of H2O = 1 x 5.02kJ/0.1 = -50.2kJ mol^-1
- 0.1moles H2O = 5.02kJ 1 mole of H2O = 1 x 5.02kJ/0.1 = -50.2kJ mol^-1
- moles of acid = c x V = 1 x 0.1 = 0.1moles => moles of H2O = 0.1moles
- mole ratio = 1:1
- Eh = cm/\T = 4.18 x 0.2 x 6`C = 5.02kJ
- vol. H2O = vol. acid + vol. alkali = 100 + 100 = 200cm3 (0.2kg - 1cm3 H20 = 1g)
- Example: HCl(aq) + NaOHkJ(aq) = NaCl(aq) = H2O(l)
- the enthalpy of neutralisation is
the energy change when an acid is
neutralised to form 1 mole of water
- Writing Standard
Enthalpy Equation
- combustion
- there must ONLY
be 1 mole in the
balanced equation
- C2H6 + 3 1/2O2 = 2CO2 + 3H2O
- C2H6 + 3 1/2O2 = 2CO2 + 3H2O
- there must ONLY
be 1 mole in the
balanced equation
- solution
- only
1
mole
- (NH4)2SO4(s) = 2NH4+(aq) + SO4 2-(aq)
- (NH4)2SO4(s) = 2NH4+(aq) + SO4 2-(aq)
- only
1
mole
- neutralisation
- 1 mole
- HCL(aq) + NaOH(aq) = NaCl(aq) + H2O(l)
- HCL(aq) + NaOH(aq) = NaCl(aq) + H2O(l)
- 1 mole
- combustion
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