Enthalpy changes Flashcards
exothermic and endothermic reactions
exo- is a reaction where heat energy is released to the surroundings
deltaH is negative
endo- is a reaction where heat energy is absorbed from surroundings
deltaH is positive
average bond enthalpy
the average enthalpy change for the breaking of 1 mole of bonds in gaseous molecules
standard conditions
Temp= 25⁰c
pressure= 1 atmosphere or 100 kPa
solutions must have conc of 1 moldm-³
⁰ means standard conditions
DeltaH⁰ for example is enthalpy change under standard conditions
enthalpy + enthalpy change
H, is a measure if heat energy of a system
not possible to measure H of a system but we can measure enthalpy change when heat is added or released from system
enthalpy change, deltaH:
is the amount of heat released or absorbed by chem reaction at constant pressure, Kj mol-¹
it involves an exchange of heat energy between system and surroundings
can be endo or exo
if it is negative, heat has been released
enthalpy of system decreases
temp of surroundings increase
exothermic enthalpy change
if positive, heat energy has been absorbed from surrounding
enthalpy of system increases
enthalpy change is endo
enthalpy profile diagram
shows change in enthalpy between products and reactants
enthalpy change in reaction = enthalpy of products- enthalpy of reactants
delta rH = Hp - Hr
exothermic-
energy is released
so products line is lower than reactants
activation energy ( Ea) is an arrow between the reactants and top of curve
enthalpy change (Delta rH) is an arrow down from the reactants to products
endothermic-
energy is absorbed
so products line is higher than reactants
Ea is still reactants to top of curve
Delta rH is arrow up from reactants to products
REMBER FOR BOTH WRITE PRODUCTS ON PRODUCTS LINE AND REACTANTS ON REACTANTS LINE INCLUDING STATE SYMBOLS
ALSO NEVER USE DOUBLE SIDED ARROWS EITHER UP OR DOWN
AND LABLE AXIS WITH ENTHALPY (Y) AND REACTION PATHWAY (X)
bond fission
this is the breaking of a chem bond
2 ways a covalent bond can be broken- homolytic and heterolytic
homolytic:
bond breaks equally
1 electron to each atom
show with half arrows -fish head arrow
makes 2 radicals (species with an unpaired electron)
Heterolytic:
bond breaks unevenly
both electrons to 1 atom
shown by full arrow
make one - and one + ion
breaking and forming bonds
breaking for covalent bonds in reactants and forming of new covalent bonds in products
breaking bond= endothermic as it requires energy ,DeltaH is positive
making bonds= exothermic as it releases energy ,DeltaH is negative
Average bond enthalpy DeltaEH⁰
the breaking of 1 mile of bonds in gaseous molecules
always positive as it is breaking bonds so is endothermic, requires energy
example
average bond enthalpy, F-F = 158 KJmol-1
F-F(g)—-> 2F(g) average bond enthalpy is +158 for F-F and -158 for 2F (as it is reverse reaction)
it is effected by large numbers so just x the average bond enthalpy by no
bond strength can also be effected depending in enviroment it is formed in
Calculate deltaHR( overall energy change)
BALANCEING NUMBERS ARE INCLUDED
DeltaRH = DeltaH breaking bonds- DeltaH forming bonds
if answer is - it is exothermic reaction
if + it is endothermic overall
Enthalpy change of neutralisation
Delta,neut,H⁰
The standard enthalpy for neutralisation is the enthalpy change that accompanies the formation of 1 mole of H2O from neutralisation under standard conditions
example:
HCl(aq) + NaOH(aq) –> NaCl(aq) + H2O(l)
try not to place balancing no infront of H2O as it is supposed to be one mole
if you have to x Delta neutH⁰ by the no
enthalpy change of formation
Delta,f,H⁰
the standard enthalpy change for formation is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions
example:
H2(g) + 1/2 O2(g) —> H2O(l)
again try not to put balancing no infront of product as it is supposed to be 1 mole but if you do × DeltafH⁰ by the no of moles
Enthalpy change of combustion
Delta,c,H⁰
the standard enthalpy change of combustion is the enthalpy change for complete combustion of 1 mole of a substance under standard conditions, all reactants and products in standard states
example:
CH4(g) + 2O2(g) —> CO2(g) + 2H2O(l)
once again try not to use balancing no on reactant that isnt O2 as supposed to 1 mole but if you do just × DeltacH⁰ by no
Hess’ law
the enthalpy change of a reaction depends only on the initial and final states and is independent of the route taken
e.g.
A deltaH1 B
deltaH2 deltaH3
c
shows 2 routes of converting A too B,from Hess’ law we know the enthalpy change of both is the same( as long as A and B are in same states) so
deltaH1 = deltaH2 + deltaH3
anticlock wise= clockwise
Formation- FliPeR both arrows point up
ER+EP= formation
Combustion- CRaP
both arrows point down
ER-EP= Combustion
enthalpy change can’t be measure directly sometimes so we use these cycle instead
example of when this happens is:
if byproduct is created in side reactions
or If the reaction can’t occur under standard conditions
example of Hess cycle for formation
use Hess cycle and apply Hess law to determine enthalpy of reaction
MgCO3 —-> MgO + CO2
DeltaFH of MgCO3= -1096 Kjmol-1
MgO= -602
CO2= -394
- write out equation again
2.write out elements that form it below - draw 2 arrows from bottom eq to top one to reactants one to products
- either Flipper or clockwise arrows= anticlockwise
1.MgCO3 —> MgO + CO2 - ^ ^
lI ll - Mg + C + 1/2 O2
- -996 - -1096= +100 Kjmol-1
sum of products - sum of reactants = enthalpy change
Example of Hess cycle for combustion
construct enthalpy cycle to determine enthalpy of formation of methane
C + 2H2 —-> CH4
enthalpy change of C:
C= -393.5 Kjmol-1
H2= -285.8 Kjmol-1
CH4= -890.3 Kjmol-1
1.rewrite out equation
2. write out CO2 and H2O below (balance them)
3.draw 2 arrows pointing down 1 from products 1 from reactants
4.Crap or Clockwise= anticlockwise
1. C + 2H2 —-> CH4
3. ll ll
^ ^
2. CO2 + 2H2O
4. - 965.1 - -890.3 = -74.8 Kjmol-1
sum of products - sum of reactants
