Reaction Kinetics
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Defn of rate of rctn units eqn
change in conc of a particular prod/rct per unit time mol dm^-3 s^-1 rate of eqn = change in conc / time taken
Defn of rate eqn eqn
Experimentally determined mathematical exp b/w the rctn rate and the conc of rcts Rate = k [A]^m [b]^n
Defn of order of rctn
the order of rctn wrt a rct is the power in its conc term in the rate eqn and is experimentally determined
Defn of rate const
Is the proportionality const in the exp-determined rate eqn. It is a const at a given temp
Defn of instantaneous & initial & ave rate
Instantaneous rate - is the rate at a particular time initial rate - the instantaneous rate at time=0 Ave rate - is the change in conc of a rct or a product over that time interval
In prac, why may the initial rate ≈ ave rate provided（2）
• time interval is small enough 2. time interval starts at t=0
Interpreting rate eqn units formula what order rctn
When rate = k ——– zero-order rctn units of k: mol dm^-3 s^-1 rate = k[A] ——- first-order rctn units of k: s^-1 rate = k[A][B] &. =k[A]^2 ——– second-order rctn units of k: mol^-1 dm^3 s^-1
Defn of first order rctn & eqn & graph
One in which the rctn rate is directly prop to the conc of a single rct rate = k[A]
Defn of 2nd order rctn & eqn & graph
One in which the rctn rate is directly prop to the sq of the conc of a single rct rate = k[A][B] & =k[A]^2
Defn of zero-order rctn & eqn & graph
One in which the rctn rate is independent of the conc of a single rct rate = k
Defn of half life how to find it (2) & working on graph The proving statement too
Time taken for the conc of a rct to decrease to half its initial value From the graph, 1st t1/2 = time taken for […] to decrease from …. to …. = t1 2nd t1/2 = time taken for […] to decrease from …. to …. = t2 Since t1 = 12, t1/2 is const 1st order wrt [A] On graph draw the & in ans: Since the half-life is approximately constant (averaget1/2 = 15.5min), the reaction is first order with respect to N2O3 ALSO t1/2 = ln2/k
Pseudo-order rctns (3) and the way of writing the k’ eqn
Presence of a large excess of a rcnt Rcnt is also the solvent Presence of a catalyst eg rate = k[S2O82−][I−] = k’[I-] where k’ = k[S2O8 2-] = (0.2)(0.5)
Finding order of rctn (3)
By inspection method By Calc By conc-time graph
Find k
sub values into rate eqn for a certain exp (any exp will do)
Why is sth used in large excess
Ensure that [Q] stays effectively const throughout the rctns so that the effect of changing [P] on the rate of the rctn can be studied
Exp-tech Physical methods • gas (1) • colour intensity (3) Chem methods • sampling followed by quenching then titrimetric method (3) • ‘clock rctns’ (2)
look at notes
Factors affecting rate of rctn (4)
?
‘Clock rctns’ why must the beaker be the same why must the total vol be const why is initial value inversely prop to the time taken for a___ solid/colour to appear
ensure depth of rctn mixture remains the same, same amt of __ needs to be precipitated before the cross is obscured ensures initial conc of each reactant in the rctn mixture is directly prop to its vol used time interval is small and measured from 0
Defn of rctn mechanism
collection of elementary steps in the proper sequence showing how rcnt particles are converted into products. It is the explanation of how a rctn takes place
Defn of elementary step
distinct step in a rctn mechanism which describes a single molecular event that involves breaking or making bonds
Defn of rate-determining step
Slowest step and highest Ea (find actual defn from tut?)
What is the molecularity of an elementary step eg unimolecular step, bimolecular step
No. of rct particles taking part in that step
Defn of an intermediate
a species that is formed in 1 step of a rctn mechanism and consumed in a subsequent step
Defn of single/multi-step rates How to find it from stoic eqn deducing rate eqn
Single step rctns is one that takes place in a single step and is termed an elementary rctn