Class 4: Kinetics and Equilibrium
Changing the Temperature in a Reaction Mixture
- Exothermic reactions release heat (- H kJ/mol) (heat is a product) - Endothermic reactions consume heat (+ H kJ/mol) (heat is a reactant) - increasing the temperature of an endothermic reaction will shift the equilibrium toward the products ex: the Haber process is an exothermic reaction. If you increase the temperature at which the reaction takes place once it's reached equilibrium, the reaction will shift to the left in order to consume the extra heat. thereby producing more reactants. - If you decrease the temperature at which the reaction takes place once it's reached equilibrium, the reaction will shift to the right in order to produce extra heat, thereby making more product. - Lowering the temperature favors the exothermic reaction - Raising the temperature favors the endothermic reaction - Changes in temperature will affect (unlike changes in pressure or concentration) the reaction's Keq Heating a reaction gets it to equilibrium faster (the rate of the reaction proceeds faster - a kinetic property)
exothermic reactions
- H, S can be - (entropy decreases) or + (entropy increases)
The researcher could NOT have discovered the rate law for the homogeneous reaction by measuring the change in total pressure in the reaction chamber if the reaction had: A. used a catalyst. B. taken place at room temperature. C. been allowed to proceed to completion. D. produced the same molar quantities of products as reactants.
- If the reaction produced the same molar quantities of products as reactants, then there would have been no change in pressure, but pressure changes are precisely what determines the reaction rate for a homogeneous reaction involving gases.
ΔH = Hproducts - Hreactants.
- In order for the reaction to be exothermic (negative ΔH), Hreactants must be greater than Hproducts. - The H of a molecule is the sum of the bond energies. Watch out, though, because you have to remember that bond strength is inversely related to bond energy! So the bond strength of the reactants have to be less than that of the products
reaction quotient (Q) (Q depends on 2 things: 1 that Keq also depends on + another that Keq doesn't)
- Q describes the distance from equilibrium that a reaction is = the ratio of the instantaneous/initial product and reactant concentrations (snapshot/moment in time of the reation) - so use when the reaction is not at equilibrium **value calculated using initial concentrations (while K is equilibrium concentrations) ***Q depends on the temperature AND the concentration (K only depended on temperature)
Factors that affect reaction kinetics (3)
- Temperature, reactant concentrations, and activation energy are the three major factors that play a role in a reaction's kinetics. - If the temperature of the beaker changes, the rate of the forward reaction will be affected. - When more reactants are added or the mixture stirred, the probability of collisions among the molecules increases. Thus, the forward reaction rate will increase. (While removing products will shift the equation to the right (favoring the production of more products) by Le Chatelier's Principle, this is an effect on the equilibrium, or thermodynamics, of the reaction and does not affect the rate of the forward reaction.)
PCl5(g) <> PCl3(g) + Cl2(g) ΔHo = 92.5 kJ/mol Two identical reaction vessels, X and Y, are charged with equal amounts of gaseous PCl5, and both trials are allowed to reach equilibrium. If flask X is held at 300 K while flask Y is maintained at 500 K, which of the following will be true about the reactions? A. Equilibrium is reached faster in flask Y because higher temperatures shift the reaction toward products. B. Equilibrium is reached faster in flask Y because ΔGY is more negative than ΔGX. C. Equilibrium is established at the same rate in both flasks because the K value of a reaction is a constant. D. Equilibrium is reached faster in flask Y due to the increased collision frequency of reactant molecules.
- The speed that equilibrium is reached is based on the kinetics of the forward and reverse reactions and not on any thermodynamic parameters. Since both the forward and reverse reactions of the equilibrium will be sped up by increasing temperature, flask Y will reach equilibrium faster (eliminate choice C). A = thermodynamic property B = thermodynamic property - The reason must be a kinetic principle, not a thermodynamic principle, so using ΔG or Le Châtelier's Principle, both thermodynamic concepts, as explanations must be wrong (eliminate choices A and B). - According to the kinetic molecular theory, temperature is a measure of the average speed (velocity) of the particles. Increasing temperature increases speed, which in turn increases the frequency of collisions between the particles and the vessel. D is correct
Le Chatelier's Principle
- a system at equilibrium will try to neutralize any imposed change (or stress) in order to reestablish equilibrium (stresses = changes in concentration, pressure, volume, and temp) - the stress cannot be entirely eliminated ONLY minimized ie. if you add more reactant to a system that is at equilibrium, the system will react by favoring the forward reaction in order to consume that reactant and reestablish equilibrium ex: N2(g) + 3H2(g) <--> 2NH3(g) + heat - adding more ammonia will create an excess of product, so by consuming some of the added ammonia, the ratio of products to reactants would decrease towards the equilibrium ratio = so the reverse reaction will be favored (the system shifts to the left), converting ammonia into nitrogen and hydrogen until equilibrium is restored - removing ammonia, Q < Keq so reaction will shift in forward direction
Adding a Catalyst
- adding a catalyst to a reaction that is already at equilibrium has no effect (as it increases the rate of the forward and reverse reactions equally) - a catalyst increases the reaction rate but does NOT affect the equilibrium
Thermodynamic Properties
- beginning state of reaction and end state of reaction for products (change in something) - G, H, S - spontaneity - stability - equilibrium (and shifting equilibrium, like removing products) - energy - free energy - intermediate
reaction rates
- can be measured in terms of the change in concentration of a reactant/product over time (rate that reactants are used up or the rate that the products are created) rR -> pP
Videos: Volume and Pressure Effects on equilibrium
- changing the pressure stresses an equilibrium if the number of moles of gas changes in the reaction - the pressure change must also be due to a volume change (P and V are inversely related) **if the volume does not change, then the equilibrium will NOT be stressed Increase Ptot by decreasing V, shifts toward: FEWER MOLES OF GAS Decrease Ptot by increasing V, shifts toward: MORE MOLES OF GAS *This could occur through the addition of an inert gas in a flexible container, like a cylinder with a moving piston. Addition of an INERT GAS WILL CHANGE THE TOTAL PRESSURE of the system ** If inert gas is added to a system of CONSTANT VOLUME IT WILL HAVE NO CHANGE IN EQUILIBRIUM ***If inert gas is added to a system of CONSTANT PRESSURE it would stress the system and equilibrium
Keq - still at equilibrium, Keq describes position at equilibrium - the ratio of equilibrium products and reactants
- describes the position of the equilibrium = the ratio of the equilibrium product and reactant concentrations *calculated using equilibrium concentrations *K is ONLY dependent on the temperature
reaction coordinate
- difference of reactant energy and product energy determines deltaH/deltaG (be sure to check axis label) - peaks = transition states (kinetic) - valleys = intermediates (thermo) - activation energy is the difference between reactant energy and highest energy transition state -- catalysts LOWER the peaks and don't change the valleys (don't change the intermediates) - if you have more than one peak, the highest peak (slowest step) is known as the rate determining or limiting step (RDS)
Activation energy (Ea)
- every chemical reaction has one = the minimum energy required of reactant molecules during a molecular collision in order for the reaction to proceed to products - if reactants possess the necessary Ea, they can reach a high-energy (and short lived, difficult to isolate) transition state (aka activated complex) - the transition state is always an energy maximum, and is therefore distinct from an intermediate (intermediates are produced in an early step of the mechanism, and are later used up so they do not appear as products of the overall reaction = thermo quantities because they do not affect the rate)
A single step mechanism
- ie where a reactant (NO2) is decomposed in a single step: 2 NO2 ( g ) <> 2 NO( g ) + O2 ( g ) - should not produce any detectable intermediates. Detection of intermediates would be most useful to challenge this hypothesis. *- the faster the elementary step, the steady state concentration of the intermediate becomes smaller and harder to detect* (None of the other three experimental approaches would address the question of whether only one or more than one step is involved in the mechanism. Changing temperature, adding a catalyst, and altering concentration of reactants are all expected to change the rate of any reaction.)
Solubility Effects in Acid/Base
- if you REMOVE carbonate (CO32-) from solution by using an acid and making carbonic acid (CH2O3) you would be able to dissolve more of the solid CaCO3 because you are getting rid of common ion and shifting the equilibrium to the right
to determine which way a reaction will shift using pressure
- increase [ ] or pressure of a reactant, will shift towards products REMEMEBR PRESSURE FOLLOWS [ ] TRENDS - increase pressure, will shift towards side with less moles - decrease pressure, will shift towards side with more moles Le Châtelier's Principle implies that increasing the pressure by decreasing the volume of a container will shift the reaction to the side with the fewer moles of gas. Since the reactant side has 1 + 3 = 4 moles of gas while the product side has just 2, the forward reaction will be favored.
Adding an Inert (or Non-Reactive) Gas
- inert gases don't participate in any reaction, so adding it (i.e He) into a constant volume reaction container (rigid container) will not change the partial pressure or the concentration of reactants and products (if neither of these values change, then there is no change in equilibrium) However, if He is injected into a constant pressure container, like one with a movable piston, the extra gas particles will push against the piston, raising it to increase the volume and equilibriate the internal pressure of the gases with external pressure. - since the volume increases, the partial pressures of the gases involved in the reversible reaction will change, causing a shift in the equilibrium as described above for volume changes (increase volume, will shift towards side with greater number of moles)
Thermodynamics and Equilibrium
- look at bottom two equations - G' = Gibbs free energy of a reaction at standard conditions - deltaG change with changing reaction composition until it reaches 0 G' < 0; Keq > 1, products are favored at equilibrium (more products) G' = 0; Keq = 0, products and reactants are present in roughly equal amounts at equilibrium G' > 0; Keq < 1, reactants are favored at equilibrium (more reactants) G = 0 then Q = K
Rate Laws
- only those that are involved in the rate-determining step (the slow step) are part of the rate law expression ex: on pg. 190 ex: in picture - in experiment 1 - 2 only [Y] changes, it doubles from 0.020 - 0.040 and the reaction rate doubles from 2.0 x 10^-4 to 4.0 x 10^-4 so the reaction rate is proportional to [Y] and x=1 (first order_ - now look at [X], [X] changes in exp. 2-3 and is doubled just like the rate = first order
Kinetic Properties
- rate - mechanism (increasing [reactants] increases forward rate - catalyst - transition state - activation energy
The equilibrium constant (Keq)
- solids and pure liquids are not included, because their concentrations don't change (s) or (l) - the value of Keq for a given reaction is a constant at a given temperature - if the temperature changes, the a reaction's Keq value will change Keq < 1 at equilibrium. reaction favors the reactants (i.e there are more reactants than products at equilibrium) Keq = 1 at equilibrium. reaction has roughly equal amounts of reactants and products Keq > 1 at equilibrium. reaction favors the products (i.e there are more products than reactants at equilibrium) *to determine size of Keq - look at ratio of products to reactants in equation. More products will yield a greater Keq value deltaG' = -RTln(Keq) - temperature can change a reaction's Keq - IF DELTAG' GOES UP, KEQ GOES DOWN
Concentration of a solution
- the concentration of a solution tells you how much solute is dissolved in the solvent - a concentrated solution has a greater amount of solute per unit volume than a solution that is dilute - a saturated solution is one in which no more solute will dissolve. At this point, we have reached the molar solubility of the solute for that particular solvent, and the reverse process of dissolution, called precipitation, occurs at the same rate as dissolving. Both the solid form and the dissolved form of the solute are said to be in dynamic equilibrium -- if a solution is saturated; the rate at which ions go into solution is equal to the rate at which they precipitate out
Solubility Product Constant (Ksp)
- the extent to which a salt will dissolve in water - Ksp is another equilibrium constant, one in which the reactants and products are just the undissolved and dissolved salts
the rate constant (k) (2 factors affect it)
- the rate constant (k) is proportional to the probability that a collision will result in a successful reaction - the rate constant is indirectly proportional to the activation energy (Ea) -- adding a catalyst lowers the activation energy - the rate constant is directly proportional to the temperature - the rate constant is NOT affected by the concentration of the reactants -- it's a constant (K) so it doesn't matter how many of the molecules you have, rather the probability that those molecules will collide
In order for a reaction to occur... (factors that affect rate)1
- the reactants must collide.... higher concentrations = faster rate higher temperature = faster rate - with proper orientation... fixing orientation (via catalyst) = faster rate (think about how you can't just throw your keys at a lock and it opens, your fingers must insert the key and turn; fingers = catalyst) - ...with a minimum energy (activation energy) higher temperature = faster rate require less catalyst = faster rate
disorder decreases when
- two molecules become one: H+(aq) + OH-(aq) --> H2O(l) - when two aqueous species turn into a pure liquid (this increases order) when disorder decreases, deltaS < 0 (1 mol - 2 mol) deltaS = nSproducts - nSreactants
Electrolytes
- when ionic substances dissolve, they dissociate into ions - free ions in a solution = electrolytes, because the solution can conduct electricity - solutes that dissociate completely (like ionic substances: NH4I, LiF, AgBr) = strong electrolytes; best conductors - solutes that remain ion-paired to some extent = weak electrolytes (i.e H2O2) - covalent compounds that don't dissociate into ions - nonelectrolytes - C6H12O6 is non-ionic, so it does not dissociate, i=1 (the van't Hoff factor for almost all biomolecules - hormones, proteins, steroids = 1) - NaCl dissociates into Na+ and Cl-, i=2 - HNO3 dissociates into H+ and NO3-, i=2 - CaCl2 dissociates into Ca2+ and 2Cl-, i=3
Videos: Temperature Effects
- you must know if reaction is endo/exothermic (if you don't know then you can't predict) endo = heat is a reactant (+deltaH kJ/mol) exo = heat is a product (-deltaH kJ/mol) increasing the temperature is the same as increasing the concentration of heat (adding a reactant or a product) decreasing the temperature is the same as decreasing the concentration of heat (removing a reactant or a product)
Reaction rate
1. the lower the Ea, the faster the reaction rate 2. the greater the concentrations of the reactants, the faster the reaction rate (favorable collisions are more likely as the concentration of reactant molecules increase) 3. the higher the temperature of the reaction mixture, the faster the reaction rate - at higher temperatures, more reactant molecules have sufficient energy to overcome the Ea barrier, and molecules collide at a higher frequency, so the reaction can proceed at a faster rate How to get more collisions: - increase temp - increase [ ] - increase pressure (if everything is in gas phase) or decrease the volume How to get proper orientation: - adding catalysts - changes ways reactants are colliding which increases specificity of RXN How to lower EA: - add catalyst - increase temp
Why do kidney stones form in some individuals treated with Compound 1? A. [Ca2+] + [C2O42-] > Ksp B. [Ca2+][C2O42-] > Ksp C. [Ca2+] + [C2O42-] < Ksp D. [Ca2+][C2O42-] < Ksp
1st. Kidney stones = solid = precipitate so must have Qsp > Ksp Calcium oxalate precipitates when [Ca2+][C2O42-] exceeds the solubility product constant Ksp when [Ca2+][C2O42-] > Ksp
LR question N2( g ) + 3 H2( g ) 2 NH3( g ) ΔH = -91.8 kJ/mol If 6 moles of H2 react with excess N2, how many moles of NH3 are produced after one pass over the catalyst in the Haber process? A. 0.1 moles B. 0.6 moles C. 2 moles D. 4 moles
6 mol H2/3mol H2 = 2 mol x 2mol of NH3 = 4 If 6 moles of H2 react with excess N2, 4 moles of ammonia will ideally be produced since the ratio of H2:NH3 is 3:2 in Equation 1. However, the passage states that after one pass the yield is 15%, eliminating answer choice D. Only 15% of 4 moles, or 0.6 moles of ammonia are expected, making choice B the correct answer.
Arrhenius equation
A = Arrhenius factor (takes into account the orientation of the colliding molecules) Ea = activation energy R = gas law constant (0.08 = 0.1) T = temp (K) solve for k k = lnA-(Ea/RT) adding a catalyst (decreasing Ea) or increasing T will increase k - rough rule of thumb = the rate (k) will increase by a factor of 2 - 4 for every 10-degree Celsius increase in temperature
Common ion effect
A salts solubility will DECREASE if it is added to a solution containing a common ion Ca3(PO4)2(s) <-> 3Ca2+(aq)+2PO43-(aq) *the solubility of calcium phosphate decreases in a solution of CaCl2 because you already have Ca2+ in solution which is going to prevent more of the Ca3(PO4)2 from being able to dissolve because it throws off the ratio of calcium to phosphate SHIFTS REACTION TO THE LEFT AND THUS SOLUBILITY DECREASES - relate to Le Chatlier's because adding more of a common ion could increase the [product] of that ion and shift it to the left (or vice versa) = decreasing its solubility
Addition of reactant to a previously equilibrated system will result in which of the following? A. K > Q and the forward reaction predominates B. K > Q and the reverse reaction predominates C. K < Q and the forward reaction predominates D. K < Q and the reverse reaction predominates
A. K > Q and the forward reaction predominates Since a K or Q expression is a ratio of [products]/[reactants], addition of a reactant will decrease the reaction quotient and have no impact on the equilibrium constant which depends only on temperature, thus K > Q (choices C and D are incorrect). When K > Q, the reaction proceeds forward in order to use reactants, make products, and reestablish equilibrium where Q = K (choice B is incorrect and A is correct).
When is the reaction quotient equal to the equilibrium constant? A. When ΔG = 0 B. When the forward reaction has stopped C. When the reverse reaction has stopped D. When product concentrations are equal to reactant concentrations
A. When ΔG = 0 K = Q A reaction has reached equilibrium once the reaction quotient equals the equilibrium constant Forward and reverse reactions do not cease at equilibrium but simply occur at the same rate, resulting in no net change in reactant or product concentrations (choices B and C are incorrect). At equilibrium, the concentration of products and reactants are not necessarily equal (choice D is incorrect).
Equilibrium
As the reaction proceeds, products begin to form and eventually build up, an some of them begin to revert to reactants - once products are formed both the forward and reverse reactions will occur - ultimately the reaction will come to an equilibrium, a state at which both the forward and reverse reactions occur at the same constant rate - at equilibrium the overall concentration of reactants and products remains the same, but at the molecular level, they are continually interconverting - because the forward and reverse processes balance one another perfectly. we don't observe any net change in the concentrations NO CHANGE IN CONCENTATIONS when a reaction is at equilibrium (and only at equilibrium), the rate of the forward reaction is equal to the rate of the reverse reaction - equilibria occur for closed systems (no new reactants, products, or other changes)
A chemist is working with a well-characterized reaction in her laboratory. Addition of an unknown compound to the reaction results in an increase in initial reaction rate. However, upon addition of larger quantities of the same compound, no further increase is detected. What is the most likely identity of the unknown compound? A. An intermediate B. A catalyst C. An inhibitor D. A product
B. A catalyst Addition of a catalyst will increase the rate of a reaction by decreasing the activation energy of the step in question but will not continue to do so with increasing concentrations of catalyst (choice B is correct). While addition of an intermediate may increase the rate of a reaction, the initial rate will not plateau at higher concentrations (choice A is wrong). An inhibitor will likely decrease the reaction rate (choice C is wrong) and addition of a product is unlikely to speed the reaction (choice D is wrong).
The ΔG° of a reaction increases following an increase in temperature. What impact would this have on the equilibrium constant? A. It increases. B. It decreases. C. It does not change. D. Inadequate information is provided to determine the outcome.
B. It decreases. If the ΔG° of a reaction increases with increasing temperature, this will cause a decrease in the equilibrium constant Qualitatively, as ΔG° increases, the reaction becomes less spontaneous and favors the reactants to a greater degree. With an increase in reactants at equilibrium, the K decreases since K is a ratio of [products]/[reactants]. Quantitatively, ΔG° = -RT ln K, thus as ΔG° increases, K must decrease.
How will the rate of a catalyzed reaction be affected if the solid catalyst is finely ground before it is added to the reaction mixture? A.The rate will be faster because a greater mass of catalyst will be present. B.The rate will be faster because a greater surface area of catalyst will be exposed. C.The rate will be slower because the fine catalyst particles will interfere with product formation. D.The rate will remain the same because the mass of catalyst will be the same.
B.The rate will be faster because a greater surface area of catalyst will be exposed. grinding a heterogeneous catalyst increases the amount of catalyst available to the reaction and therefore increases its rate.
The equilibrium BaCrO4(s) <> Ba2+(aq) + CrO42-(aq) exists in a saturated aqueous solution of BaCrO4. Dissolution of Na2CrO4 in a saturated aqueous BaCrO4solution would: A.have no effect on the position of this equilibrium. B.shift this equilibrium left. C.shift this equilibrium right. D.shift this equilibrium first right and then left.
B.shift this equilibrium left. because dissolution of Na2CrO4 would introduce the common ion, CrO42-, which would reduce the solubility of BaCrO4 due to the common ion effect.
Solubility Equilibria (Ksp vs. S)
Compounds that dissociate into ions when dissolved in water can be discussed in terms of ... - Solubility Product (Ksp) = the equilibrium constant for the dissociation equation; used for ionic compounds with low solubility (like Ca3(PO4)2 not NaCl) - Molar Solubility (S) = the amount of a substance that can dissolve in a specific solvent at a specific temperature (3 tsp sugar into cup of water) Ksp = [3S]^3 [2S]^2 = (27S^3)(4S^2) = 1085S^5 Always soluble? Usually insoluble?
Videos: Concentration/Partial Pressure Effects
Concentration and partial pressure changes behave the same - increase [ ] or Pn of a reactant, shift toward product - decrease [ ] or Pn of a reactant, shirt toward reactant - increase [ ] or Pn of a product, shift toward reactant - decrease [ ] or Pn of a product, shift toward product *solids/solvents do not affect the position of the equilibrium (because they are not involved in equilibrium expression) **inert gases do not affect the position of the equilibrium (partial pressures will remain the same no matter how much inert gas you add to system)
Changing the Volume of the Reaction Container (increase V, decrease P)
Consider a gaseous reaction (at equilibrium) with unequal numbers of moles of gas of reactants and products. - if the volume is reduced, increasing the pressure, a net reaction occurs favoring the side with the smaller total number of moles of gas - if the volume is expanded, decreasing the pressure, a net reaction occurs favoring the side with the greater total number of moles of gas (this is only true for reactions involving gases) if a reaction vessel is rigid = this means volume is constant *if the same # moles for products and reactants you wouldn't see any changes*
Enzymes alter the rate of chemical reactions by all of the following methods EXCEPT: A.co-localizing substrates. B.altering local pH. C.altering substrate shape. D.altering substrate primary structure.
D.altering substrate primary structure The primary structure of a protein substrate is the amino acid sequence of the protein. Enzymes cannot alter primary structures of protein, but can colocalize substrates, alter local pH, and alter substrate shape.
multiple equilibria
If ONE reaction in a system of multiple equilibria shifts, ALL reactions in the system shifts in the SAME direction. - think of bicarbonate buffer equation - all equilibriums will shift left, increases pH
Formation Equilibria
K > 1 favors products - the equilibrium of a coordination complex (i.e. Hemoglobin) is a formation constant: Kf - reversing an equilibrium inverts the equilibrium constant (take reciprocal: 1/Kf); DISSOCIATION CONSTANT IS THE INVERSE OF KF: (1/Kd)=Kf; Kd=(1/Kf) - combining multiple equilibria results in the multiplication of the equilibrium constants (K1 x K2 or K1 x 1/K2 for reversed 2nd reaction) *immediately notice that K1 > K2; this tells us that Cu2+ is going to have the higher affinity or tend to form the complex more than Cd2+ will **since we combined reactions you must combine the K's
reaction quotient (Q) and Keq
Q < Keq NOT in equilibrium the reaction will proceed in the forward direction to increase Q to the Keq value (in order to increase the concentration of the products and decrease the concentration of the reactants) - means there is excess reactant or shortage of product (will favor forward direction) Q = Keq reaction is at equilibrium, Keq is the condition the reaction will try to achieve G=0 Q > Keq NOT in equilibrium the reaction will proceed in the reverse direction to reduce Q to Keq (in order to increase the concentrations of the reactants and decrease the concentrations of the products) - means there is excess product or lack of reactant (shifts towards reactants)
Ion Product (Qsp) Qsp = think concentration Ksp = think equilibrium
Q = a momentary snapshot of what is happening in the reaction = the reaction quotient for a solubility reaction, unlike Ksp, the concentrations don't have to be those at equilibrium - Ksp is equal to the product of the concentration of the ions in solution when the solution is saturated (at equlibrium) Qsp < Ksp more salt can be dissolved, shifts towards products CONCENTRATION < SOLUBILITY; no precipitate; more solute can dissolve Qsp = Ksp solution is saturated, in equilibrium (rate at which ions go into solution is equal to the rate at which they precipitate out) CONCENTRATION = SOLUBILITY; no precipitate, no more solute can dissolve Qsp > Ksp excess salt will precipitate (reverse reaction would be favored) CONCENTRATION > SOLUBILITY; no more solute can dissolve
zero order rate law
Rate = k unitless the concentration will increase (x-axis) but the rate will not change - saying that the concentration has no effect on the rate
1st order rate law
Rate = k[A] (k=1/s) - the rate will increase by the same amount that the concentration increases
A warm saturated solution of KNO3 was cooled until crystals formed. The solution was then warmed at exactly the same rate until it reached the original temperature. It was observed that crystals redissolved during warming much more slowly than they formed during cooling. Which one of the following statements provides the best explanation for this observation? A. Kinetic factors are different during crystallization than during dissolution. B. The Ksp of KNO3 is constant with increasing temperature. C. KNO3(s) is much more stable than KNO3(aq). D. ΔHcrystallization is more negative than ΔHdissolution.
Rate is a kinetic principle and can only be affected by other kinetic factors. Choices B, C, and D are all thermodynamic factors that have no influence on the rate of a process. Therefore, choice A is the only possible answer.
2nd order rate law
Rate=k[A]^2 k = 1/ms - The order can be determined two ways. First, it can be noticed that doubling the concentration causes the rate to quadruple. Second, the parabolic shape of the curve shows a squared relationship.
third order rate law
Rate=k[A]^2[B] k = 1/m^2s
The addition of a catalyst to a chemical reaction will bring about:. A. an increase in activation energy and an increase in the rate of the reverse reaction. B. a decrease in activation energy and an increase in the rate of the forward reaction. C. an increase in activation energy and a decrease in the rate of the forward reaction. D. a decrease in activation energy and a decrease in the rate of the reverse reaction.
The addition of a catalyst to a chemical reaction will bring about a decrease in activation energy and an increase in the rate of the forward reaction. A catalyst increases the rate of both the forward and reverse reactions, and it does this by lowering the activation energy
2 SO2(g) + O2(g) 2 SO3(g) A chemist adds a reagent to the following equilibrated reaction and immediately observes the forward and reverse reaction rates to increase by 2.14 and 2.29, respectively. Which of the following was most likely added to the reaction? A. SO3(g) B. V2O5(s) C. SO2(g) D. O2(g)
The situation here describes the addition of a catalyst to an equilibrated reaction. Catalysts increase the rates of both the forward and reverse reaction by decreasing the activation energy. Addition of a reactant (SO2(g) and O2(g)) would initially increase the rate of the forward reaction alone while adding a product (SO3(g)) would initially increase only the rate of the reverse reaction. Thus V2O5(s) must serve as a catalyst in this reaction and results in the observed rate changes.
Rate-Determining Step (eq for reaction rate with products and reactants; two fractions)
The slowest step in a process determines the overall reaction rate - reaction rates ca be measured in terms of the change in concertation of a reactant or product over a change in time aA > pP rate = (-)(1/a)/delta[A]/deltat reaction rate is dependent on 3 things: 1. how frequently the reactant molecules collide 2. the orientation of the colliding molecules 3. their energy (minimum energy = activation energy)
Solutions
a solution forms when one substance dissolves into another, forming a homogeneous mixture - the process of dissolving = dissolution (i.e sugar dissolved into iced tea = solution) - solvation = when solvent molecules surround the solute molecules; if the solvent is water, this process is called hydration Solutions can involve any of the 3 phases of matter: - solution of 2 gases - solution of a gas in a liquid (i.e. seltzer water) - solution of a solid in a liquid (i.e NaCl salt) - solution of a solid in a solid (an alloy) - when a solution has water as the solvent = aqueous solution How do we know which solutes are soluble in which solvents? - like dissolves like - solutes will dissolve best in solvents where the intermolecular forces (IMFs) being broke in the solute are being replaced by equal (or stronger) IMFs between the solvent and the solute (think back to o-chem with the solubility rules; 5:1, charges)
Equilibrium Constants
are constant as long as the temperature is constant (i.e 25C at RT) Kc Kp Kf - FORMATION - COORDINATION COMPLEX (RATES k^-1, k) Ksp Ka Kb
Reaction Mechanism
dirty dish --> clean-and-dry-dish when in reality the reaction happens in several elementary steps: 1. dirty dish --> soapy dish 2. soapy dish --> rinsed dish 3. rinsed dish --> clean-and-dry-dish intermediates = soapy dish and rinsed dish - each are necessary for the conversion of dirty dishes to clean-and-dry-dishes, but do not appear in the starting material or the products - if you add up all of the reactants and products, the intermediates cancel out the intermediate - will be produced in one step and consumed in the next - the faster the elementary step, the steady state concentration of the intermediate becomes smaller and harder to detect
reaction mechanism (enzyme and intermediates)
enzyme/catalyst = used up in the first step but regenerated at end of reaction - [catalyst] = constant intermediate = produced in the first step and used up in the next step - [intermediate] = 0 - is actually made and used - detectable/isolable **transition states are not usually shown in mechanisms - not detectable/isolable
rate constant units
first order : s^-1 second order : M^-1s^-1 third order: M^-2s^-1 - a rate law can only be determined from experimental data or if given a mechanism, and has the general form: Rate = K[reactants]^x, where x is the order of the reaction with respect to a given reactant, and K is the rate constant - the overall order of a reaction is the sum of all the exponents in the rate law - the value of the rate constant, k, depends on temperature and Ea, and its units will vary depending on the reaction order - coefficients of the reactants in the rate limiting step of a mechanism can be used to determine the order of a reaction in the rate law; coefficients from the overall reaction alone CANNOT be used to find the order of a reaction (i.e 2A + 3B --> C + 3D; unless this reaction is the rate-determining, elementary step, we have no way of knowing what the rate law is)
Free Energy and the Equilibrium Constant (K) equation
large K = reaction favors products = smaller G - High T = High G = low Keq
Rate Laws (from a mechanism)
rate laws give the rate in terms of the initial concentrations of REACTANTS and the rate constant for the process **the rate law is ONLY written from "elementary" steps ***solids (s) and solvents (l) are not included in rate law ****the coefficients become the exponents and tell you the order of the reaction The rate law of a process is determined by the RDS
Solubility
refers to the amount of solute that will saturate a particular solvent Phase solubility rules: 1. the solubility of solids in liquids tends to increase with increasing temperature 2. the solubility of gases in liquids tends to decrease with increasing temperature (gases like lower temps) 3. the solubility of gases in liquids tends to increase with increasing pressure (Solubility = kP: P^n[ reactants]/P^n[products])) Salt solubility rules: 1. All Group I (Li+, Na+, K+, Rb, Cs+) and ammonium (NH4+) salts are soluble 2. All nitrate (NO3-), perchlorate (ClO4-), and acetate (C2H3O2-) salts are soluble 3. All silver (Ag+), lead (Pb2+/Pb4+) and mercury (Hg22+/Hg2+) salts are INSOLUBLE, except for their nitrates, perchlorates, and acetates
Endothermic reaction: the bonds in the reactants are stronger/weaker than the bonds in the products?
the bonds in the reactants are stronger than the bonds in the products H BDE= bonds broken (reactants) - bonds formed (products) *bond strength is inversely proportional to bond energy* - so stronger bonds in the reactants = less energy in the reactants - so H = products - reactants (small number) = ++H The ΔH of a reaction is determined by adding how much energy is required to break reactant bonds to the amount of energy released when product bonds are formed. In this situation, the amount needed to break bonds must have been higher, PRODUCTS HAVE MORE ENERGY = +H ON GRAPH SO BDE= BondsBroken - BondsFormed = +H so broken had more energy
As the temperature at which a reaction takes place is increased: A. the reaction rate will remain constant, but the rate constant will increase. B. the reaction rate and the rate constant will remain constant. C. the reaction rate will increase, but the rate constant will remain constant. D. the reaction rate and the rate constant will increase
the reaction rate and the rate constant will increase. An increase in the rate constant always corresponds to a faster reaction rate, so the statements "the reaction rate will increase, but the rate constant will remain constant" and "the reaction rate will remain constant, but the rate constant will increase" are eliminated. Increasing the temperature of a reaction always increases the reaction rate
A reaction between two species is experimentally observed to be second order overall. If the concentration of one of the species doubles, what happens to the reaction rate? A. The rate doubles B. The rate is halved C. Cannot be determined D. The rate quadruples
the reaction rate cannot be determined. There are three possibilities for a two species reaction to produce an overall second order rate law: (1) rate = k[A][B], (2) rate = k[A]2, (3) rate = k[B]2. Let's assume that A is the species doubled. In the first case, doubling A will cause the rate to double. In the second case, doubling A will quadruple the rate. Finally, in the third case, doubling A will cause no change in the rate. Therefore, more information is needed to answer the question properly.
endothermic reactions
+H, S can be + (entropy increases) or - (entropy decreases) - these reactions will absorb the most heat upon dissolving - a salt that absorbs the most heat upon mixing with water will make the best icepack (choose answer with the greatest ++H)
to determine which way a reaction will shift using enthalpy
+H, heat is a reactant (endothermic), so increasing the temperature will cause it to shift to the right
Catalysts
- - provide reactants with a different route, usually a shortcut, to get to products - will almost always make a reaction go faster by either speeding up the rate-determining step or by providing an optimized route to products - lowers the Ea of the rate-determining step and therefore lowers the highest-energy transition state - catalysts decrease the Ea of the forward reaction and reverse reaction (this is why they don't have a net effect on system that's already at equilibrium) - CATALYSTS INCREASE THE RATES OF BOTH THE FORWARD AND REVERSE REACTION BY LOWERING THE EA reactants vs. catalysts: - reactants are converted to products but a catalyst remains unchanged at the end of a reaction - a catalyst may undergo a temporary change during a reaction, but it is always converted back to its original state - like intermediates, catalysts are not included in the overall reaction equation A catalyst does not change any of the thermodynamic quantities: - G, H, S of a reaction - the heat of the reaction (delta H; enthalpy) will be the same with or without a catalyst (**think about H arrow: reactants to products, doesn't change)
Why is nitrogen gas normally unreactive? A. N2 is polar, and does not react with non-polar molecules. B. N2 is present only in a small percentage in the atmosphere. C. N2 is a gas at atmospheric pressure and room temperature. D. N2 has a triple bond with high bond dissociation energy.
- D. N2 has a triple bond with high bond dissociation energy - Nitrogen gas is non-polar, eliminating choice A. The passage states that nitrogen gas composes approximately 80% of dry air, eliminating choice B. Although N2 is a gas at standard state, that does not explain its reactivity. Choice D is a better answer because it explains why N2 is unlikely to break its bonds and form new bonds with other atoms.
