3.1.8 thermodynamics Flashcards

Question

enthalpy of solution

Answer 1

enthalpy change when one mole of an ionic solid dissolves in an amount of water large enough so that the dissolved ions are well separated and do not interact with each other solid ionic lattice --->>> free ions in solution VARIES - will explain

Answer 2

solid ionic lattice ↓ ↓[[endothermic; energy needed to break ionic bonds]] ↓ ionic bonds between oppositely charged ions are broken ↓ ↓[[exothermic; forming FoA between h2o and ions]] ↓ ions become surrounded by water molecules (polar solvent) ↓ free ions in solution

Answer 3

delta -ve O is attracted to cation. so the water surrounds them, Os facing the cation. delta +ve H is attracted to anion. so the water surrounds them, Hs facing the anion forms hydration shells around the ions - NOT a complex, water molecules orienting themselves based on eFoA

Answer 4

enthalpy change when one mole of gaseous ions become hydrated (dissolved in water) exothermic gas ions -> dissolved ions

Answer 5

gas ions ↓(1) ↓(2) ionic solid →(3)→ dissolved ions 1- lattice enthalpy of formation 2- hydration enthalpies 3- enthalpy of solution

Answer 6

1. solid ionic compound 2. arrowing going up (endo) showing lattice enthalpy of dissociation 3. goes to the ions (eg Na+ and Cl- (g/g)) 4. enthalpy of hydration going down for one of them, changes sign eg Na+ (g) to Na+ (aq) 5. enthalpy of hydration for the other, going down again (so now its Cl-(aq) and Cl- (aq)) 6. connected by arrow going up from solid ionic compound to the aq ions, endothermic, representing the enthalpy of solution

Answer 7

depending on how endothermic the LED is (if LEF remember must switch sign) and how exothermic the sum of the hydration enthalpies the enthalpy of solution could be +ve or -ve, endo or exo if LED > sum of hydr enthalpies, then endo (+) if LED < sum of hydr enthalpies, then exo (-)

Answer 8

requires no energy input - most exo reactions; a reaction that is exothermic will result in products that are more thermodynamically stable than the reactants. this causes reactions to be spontaneous (occur without any external influence). - some exo reactions require input of heat energy eg combustion, thermal decomposition - endo reactions can also be spontaneous.

Answer 9

enthalpy change of a reaction is insufficient to explain feasibility of reactions: enthalpy - system's internal energy entropy - measure of a system's disorder this explains why some endo reactions are spontaneous, because if the reaction increases entropy, will be spontaneous (probs)

Answer 10

J K^-1 mol^-1

Answer 11

systems tend towards higher entropy this means that, reactions with increase in entropy are favourable

Answer 12

sum of absolute entropy of products - sum of absolute entropy of reactants entropy values will always be positive (except for crystals at 0K; 0). but the CHANGE IN entropy values can be negative, ie a decrease in entropy if its +ve then disorder has increased if its -ve then disorder has decreased

Answer 13

bond enthalpy calcs ∆H = sum of energy in bonds broken - sum of energy in bonds formed bonds broken: R bond enthalpies bonds formed: P bond enthalpies

Answer 14

much more disordered solids < liquids < gases in terms of entropy because of this the same substance has different values of entropy depending on its state. the temperature at which there is a change in state is the temperature at which the reaction (of melting, of boiling) is feasible because change in gibbs free energy becomes 0

Answer 15

SOLID heat energy from the source causing increase in temp causes the particles to have more kinetic energy. they are vibrating more, so have more entropy. when the vibrations are enough to break bonds/forces to become liquid, LIQUID it happens. there is a jump in entropy. the period of changing state, there is no increase in temp because heat energy --> overcoming bonds, not to kinetic energy. upon continued heating when forces/bonds(?) are overcome again, GAS conversion to gaseous state. much more highly disordered.

Answer 16

temp on y, time on x - at the points of melting and boiling, the graph is horizontal. because over time (as its reached the points) energy from the heat energy input is being used to overcome bonds/forces. so temp doesnt increase over time. - then line continues going up because temp is increasing (temp as measure of vibrations of particles, so heat energy converted to kinetic energy until enough to overcome bonds/forces) ------- entropy on y, temp on x - more important, more a level - at the points of melting and boiling, graph line is vertical because entropy rapidly increases at the temp. the point at x of the vertical lines is the melting/boiling point - the vertical line corresponding to boiling point is longer than melting, because the jump in entropy (to higher entropy; more disorder) is more, because gases are extremely highly disordered

Answer 17

- look at the states, and number of particles - look at the number of particles in each state - do these for each side of the equation - remember aq means dissolves, dissociated, eg NaCl (aq) isnt one particle its two - a decrease in the number of molecules reduces entropy - a change in state will inc/dec entropy - producing a gas will increase entropy, and vv

Answer 18

change in state > number of molecules for determining the entropy change. the decrease in disorder due to less gas moles is more then the increase in disorder from more particles formed

Answer 19

if same number of molecules, same states then the change in entropy will be negligible, but we dk if there is a decrease or increase without the absolute entropy values

Answer 20

LIKELY INCREASE - change in state from s --> l --> g - increase in no. molecules LIKELY DECREASE - change in state from g --> l --> s - decrease in no. molecules

Answer 21

in the change in gibbs free energy gibbs is the thermodynamic potential of a system to do work (?) the change in gibbs free energy determines feasibility (_can_ it occur) of a reaction ∆G = ∆H-T∆S

Answer 22

- ∆G in kJmol^-1 - ∆H in kJmol^-1. - T in Kelvin, add 273 to degrees celcius - ∆S in J K^-1 mol^-1. to convert to kJ, divide by 1000 ----- for G and H, they might give you in J, which you don't need to change if they don't specify what they want answer to be in. all either needs to be kJ or J

Answer 23

driving force of all chemical processes. it must be equal to or less than 0 (so 0 or -ve value) for a reaction to be feasible so for any spontaneous reaction, ∆G = 0 or a -ve value, which is why some endo reactions are spontaneous

Answer 24

entropy value depends on state. for a given substance, entropy value changes based on state and state is determined by temp ----- it shows that temp can be calculated for when a reaction could take place, we can determine the temp at which a reaction would become feasible. reactions that arent feasible at a certain temp could/will be at another

Answer 25

changes of state are controlled by this eq, as all reactions are. SOLID. remains solid until ∆G=0, so when T is high enough (melting point) so that its 0. the reaction then becomes feasible LIQUID. remains liquid until ∆G=0, so when T is high enough (boiling point) so that its 0. the reaction then becomes feasible GAS

Answer 26

if given (the means to calculate) enthalpy change of reaction and entropy change then make ∆G=0 and rearrange for T to be subject make sure units calculate for T

Answer 27

feasibility depends on temp, and is based on gibbs. it is whether it can occur spontaneity is whether it _does_ take place or not. a reaction that is thermodynamically feasible (or thermodynamically unstable, both of those to say that change in gibbs=0 or -ve) may be KINETICALLY stable, meaning the Ea is too high and reaction does not occur. even though it could

Answer 28

∆G = ∆H-T∆S when change in gibbs will always be 0 or -ve so if ∆H is negative AND ∆S is positive because this means T∆S will always be a +ve number, and an already -ve number for ∆H - a positive, (eg -(+)3) will only make it more -ve. so ∆G will def be negative

Answer 29

∆G = ∆H-T∆S when change in gibbs will always be +ve so if ∆H is positive AND ∆S is negative because this means T∆S will always be -ve (- x + = -). and a positive - -value, --> positive + value = positive number always

Answer 30

∆G = ∆H-T∆S only if the -ve value of ∆H is MORE OR EQUAL TO T∆S so the -ve value for change in entropy needs to be smaller than -ve ∆H, so that when its times by T, this product, when added (because -(- = +) to the -ve value of ∆H, keeps it negative THEREFORE the temperature must be LOW

Answer 31

∆G = ∆H-T∆S only if the product of T∆S is BIGGER OR EQUAL TO the +ve ∆H because then the positive value of the product, when taken away from the less positive (smaller) value of ∆H, will result in the ∆G being 0 or negative because its eg 2 - +5; a more positive number being taken away (T∆S) from a smaller positive number (∆H) THEREFORE the temp must be HIGH