Pseudo first order reaction: it is a second order reaction which is made to behave like a first order reaction.
- Occurs when one of the reactants is present in excess relative to the other. E.g. hydrolysis of an ester catalyzed by hydroxyl ions in presence of large excess of alkali.
Reasons for pseudo first order reactions :
1-One or more of the reactants is present in excess that its conc. Shows no measurable change during the reaction. E.g, hydrolysis of sucrose in water
2-If the reaction has several stepsand one is slower than the rest (rate limiting step).
3-If one of the reactants is acting as a catalyst and its conc. Will not change as the reaction proceeds e.g. OH-/ H+ions for hydrolysis. E.g. hydrolysis of procaine HClin NH4Cl/ HClbuffer obeys pseudo first order kinetics.
The pseudo first order rate constant will be equal to = K x (OH-) or K x (H+) whichever maintained constant throughout the reaction.
Second order reactions :
- Reactions in which the rate depends on the concentration of two reactants or the second
power of the concentration of one reactant. E.g. saponification of ethyl acetate
- Reactions in which the rate depends on the concentration of two reactants or the second
power of the concentration of one reactant. E.g. saponification of ethyl acetate
- When A and b are two reactants (A + B ….> product)
- And the rate of decomposition of each are equal and is proportional to the product of both concentrations, then:
- -dA/dt = -dB/dt = K (A) (B)
- If (a) and (b) are the initial concentrations of A &B and (x) is the reacted (consumed) concentration of each species at time (t).
- Then the rate law can be given by:
- The rate of reaction : dx/dt = k (a-x) (b-x) ……………..eq.1
- Where (a-x) and (b-x) are the …………..
- When A and B have equal concentrations, then (a= b) :
- dx/dt = k (a-x)2 …………………………….eq. 2
- By integration: 1/(a-x) = kt + constant …………………eq.3
Second order reaction, where a=b:
- The constant can be determined by substituting x= 0 when t = 0, then the constant will be = (1/a), re-typing equation 3:
- 1/(a-x) = kt+ 1/a
- by rearrangement: Kt = x/a(a-x) or k = (1/at) (x/a-x) When x/a(a-x) was plotted against time (t) a straight line is obtained indicating a second order reaction with slope = k When x= ½ a, then: t1/2= 1/ak
- Units of K for 2ndorder reaction = liter. mol-1sec.-1