Energetics I & II Flashcards
What is the standard enthalpy change of combustion?
the enthalpy change when one mole of a substance burns completely in oxygen under standard conditions (298K, 100kPa), with all reactants and products in their standard states
What is the firm rule when talking about energy changes?
Energy can be stored and transferred, but never ‘created’ or destroyed
What is the definition of enthalpy?
the change in heat under at constant pressure.
What is the standard enthalpy change of formation?
the enthalpy change when one mole of a substance forms from its constituent elements under standard conditions (298K, 100kPa), with all reactants and products in their standard states
What is the standard enthalpy change of neutralisation?
the enthalpy change when one mole of water is produced in a neutralisation reaction between an acid and an alkali under standard conditions (298K, 100kPa). [note: not standard states because will be aqueous]
What is the standard enthalpy change of reaction?
the enthalpy change when the number of moles of substances shown by an equation as written react under standard conditions (298K, 100kPa), with all reactants and products in the states given.
What is Hess’ Law?
the enthalpy change for a given reaction is the same, independent of the route by which this is achieved (provided the temperatures, pressures, and physical states of products and reactants are the same).
(If you can’t go by A –> B, go by C)
What is bond enthalpy?
the energy required to break ONE MOLE of bonds, in the GASEOUS STATE.
What is mean bond enthalpy?
the energy required to break one mole of bonds, in the gaseous state, AVERAGED ACROSS A RANGE OF COMPOUNDS.
What is the standard enthalpy of atomisation of an element?
the amount of energy
required to produce one mole of gaseous atoms from an element in its standard state under standard conditions.
FOR EXAMPLE: Na(s) -> Na(g)
What is lattice enthalpy of formation?
the enthalpy change when one mole of an ionic compound is formed from its constituent gaseous ions.
(hence LATTICE)
(it will always be exothermic because of MAKING bonds)
FOR EXAMPLE: Ca2+(g) + Cl2-(g) -> CaCl2(s)
What is lattice enthalpy of dissociation?
the enthalpy change when one mole of an ionic compound is split up into its constituent gaseous ions.
(hence LATTICE)
(it will always be endothermic)
FOE EXAMPLE: CaCl2(s) -> Ca2+(g) + Cl2-(g)
What is first electron affinity?
the enthalpy change when one mole of electrons is added to one mole of atoms in the gaseous state to form one mole of gaseous 1- ions.
(this is like the opposition of ionisation)
FOR EXAMPLE: Cl(g) + e- -> Cl-(g)
What is second electron affinity?
the enthalpy change when one mole of electrons is added to one mole of 1- ions in the gaseous state to form one mole of gaseous 2- ions.
What is the enthalpy of hydration?
the enthalpy change when one mole of gaseous ions dissolves in water to form one mole of aqueous ions.
FOR EXAMPLE: Ca2+(g) + aq -> Ca2+(aq)
What is the standard enthalpy change of solution?
the enthalpy change when one mole of substance in its standard state is dissolved in water to form a solution where the ions are far enough apart not to interact, under standard conditions.
FOR EXAMPLE: NaCl(s) + aq -> Na+ (aq) + Cl- (aq)
What is happening to the individual chemical species step-by-step in this reaction:
Na(s) + 1⁄2 Cl2(g) -> NaCl(s)
- atomisation of Na (s)
- first ionisation of Na (g)
- atomisation of 1/2 Cl2 (g)
- first electron affinity of Cl (g)
- so now we have Na+ (g) and Cl- (g)
- lattice enthalpy of formation to form NaCl2 (s)
What is entropy
chaos/disorder
Draw a Born-Haber cycle for the formation of KCl from potassium and chlroine using the values below
atomisation of K = +121
atomisation of chlorine = +90
1st IE of K = +418
1st EA of Cl = -364
lattice formation = -710
-445 kJ mol-
Draw a Born-Haber cycle for the lattice formation of CaO from calcium and oxygen using the values below
formation of CaO = -635
atomisation of Ca = +193
atomisation of O = +248
1st IE of Ca = +590
2nd IE of Ca = +1150
1st EA of O = -142
2nd EA of O = +844
-3518 kJ mol-
Why is it that for ANY ELEMENT that the first electron affinity is negative but the second electron affinity is positive
positive = endothermic
trying to stick a negative electron onto an already negative thing
so requires energy to overcome to repulsion
What is polarising power
The ability for something to distort the electron clouds around something that is polarisable
Which one of the below has the greatest lattice enthalpy ?
- NaCl
- MgO
- BaO
MgO
talk about size and charge -> charge density
- smaller ionic radius due to its 2+ ionic charge
- has less shielding than barium
- so larger charge density
What do we assume when creating a Born-haber cycle
that the compound is purely ionic
In the data booklet for the lattice enthalpy of a coumpound and its -3245 kJ mol-.
using a Born-Haber cycle, you calculated -2884 kJ mol-. Why is there such a big difference?
We assume that the compound in the Born-Haber cycle is purely ionic and has no covalent character.
Covalent is stronger than ionic and so would require more energy to break the bonds in this compound.
Whats the difference between a theoretical value and a experimental value?
theoretical is from a data booklet/Born-Haber cycle
experimental is what actually happens
What is entropy
Disorder or chaos
Would the reaction below have a positive or negative entropy change?
C3H8 (g) + 5O2 (g) -> 3CO2 (g) + 4H2O (l)
Positive
more moles of gases on the right
gases mean the compounds are in more disorder
There is a mole increase of +2, meaning more moles in a volume and therefore more disorder.
Would the reaction below have a positive or negative entropy change?
H2O (g) -> H2O (l)
negative
going from a gas to a liquid
liquid has more order than gases, which have more disorder
What happens to entropy as enthalpy increase
Entropy increases
How does the entropy change differ from going from a solid to a liquid compared to going from a liquid to a gas
going from a liquid to a gas has a greater entropy change as there is more disorder
the particles are moving around more randomly and faster
and they are more spread out
How is dissolving ionic compounds endothermic?
NaCl(s) -> Na+ (aq) + Cl-(aq)
The H2O molecules get locked around the ion
so H2O molecules are less free to move around
so the entropy of the system decreases
What is the equation for the ΔS total
ΔS system + ΔS surroundings = ΔS total
What is the equation for ΔS surroundings
-HΔ / T
If a system increases in entropy, what happens to the surroundings usually?
the surroundings decrease in entropy
If a system decreases in entropy, what usually happens to the surroundings?
the surroundings increase in entropy
What is Gibbs a measure of
Free energy and how feasible a reaction is
How do we tell is a reaction is feasible or not using Gibbs
If Gibbs is > 0, the reaction is not feasible
If Gibbs is < 0, the reaction is always feasible
If Gibbs is 0, the reaction is JUST feasible
What is the equation for Gibbs free energy
ΔG = ΔH - TΔS System
What is the equation for ΔS System
ΔS products - ΔS reactants
Why can a reaction be feasible in terms of Gibbs, but doesn’t happen?
due to kinetic factors
- usually due to a high activation energy
- the mixture of reactants may be thermodynamically unstable (Gibbs) while being kinetically stable (Ea)
What must ΔG for a reaction to be feasible
Negative
What happens if both the ΔS System and ΔH are negative
ΔG = ΔH - TΔS System
The reaction is feasible IF the ΔH > ΔS
What happens if both the ΔS systems and ΔH are positive
ΔG = ΔH - TΔS System
The reaction is feasible IF ΔS > ΔH
What happens if ΔS system is positive and the ΔH is negative
ΔG = ΔH - TΔS System
the reaction is always feasible
What happens if ΔS system is negative and the ΔH is positive
ΔG = ΔH - TΔS System
the reaction is never feasible
what is the link between how feasible a reaction is and the position of equilibrium
the more feasible a reaction is, the more an equilibrium will move that direction
If the forwards reaction in an equilibrium is more feasible, what will happen to the value of K
Gets bigger
If the backwards reaction in an equilibrium is more feasible, what will happen to the value of K?
Gets smaller
How can we calculate Gibbs from using the equilibrium constant
ΔG = -RT lnK
How can we calculate the equilibrium constant from Gibbs
e ^ (-ΔG / RT)
Find the ΔS system for lead melting at 600K, given that the ΔH for this chnage is +5.10kJmol-. At this temperature for this change, ΔS total = 0
8.5 x 10^-3 kJ K- mol-
Find the ΔS system for lead boiling at 2017K, given that the ΔH for this change is +177kJ mol-. At this temperature for this change, ΔS total = 0.
0.0878 kJ K- mol-
explain the differences between the two ΔS system values for a substance boiling and melting, where the value for boiling is much higher than the value for boiling
Boiling means state change form liquid to gas.
Gas has more disorder due to more space and more energy, so more disorder and chaos.
Determine the temperature in degrees K, above which the following reaction will occur, given that the reaction has a ΔS value = 160.5 J K-1 mol-1
ΔH = 178 kJ mol-1
rearrange Gibbs to find T
In this case, Δ G = 0 as the we are looking for when the reaction is JUST feasible… so
0 = Δ H - TΔS
T = ΔH/ΔS
convert kJ into J
answer is 1109.7 K
Hydrogen can also be obtained by reaction of carbon monoxide with steam.
Co(g) + H2O(g) –> CO2(g) + H2(g)
ΔH = -41 kJ mol-
ΔS = - 42 J K-1 mol-1
Explain, using a calculation why this reaction should NOT occur at 1300K
Find ΔG using the normal equation
ΔG = -41 - (1300 x -0.042)
= +13.6 kJ mol-
ΔG is positive, so the reaction is not feasible at 1300K
Hydrogen can also be obtained by reaction of carbon monoxide with steam.
Co(g) + H2O(g) –> CO2(g) + H2(g)
ΔH = -41 kJ mol-
ΔS = - 42 J K-1 mol-1
The reaction is carried out at 1299K, and this is NOT feasible. Explain how this can be changed to allow this reaction to take place using a calculation
Change the temperature so it is at a point where the reaction is feasible, so we want to know when this reaction will become just feasible, so ΔG = 0
so 0 = ΔH - TΔS
T = ΔH/ΔS
-41/ -0.042 = 976.2 K
so the reaction is feasible BELOW 976.2 K
