Reaction Kinetics Flashcards
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