In this case
k = x min−1 and [A0]
= 0.01M = 1 × 10−2M
t = 1 hour = 60 min
[A]= 1 × 10−2 ( e−60x)
k = A e –(Ea/RT)
ln k = ln A – (Ea/R)
(1/T)
this equation is in the
form of a straight line equatoion
y = c + m x
a plot of lnk vs 1/T is
a straight line with negative slope
For a first order reaction
t1/2 = 0.693/k
t1/2 does not depend on the initial concentration and
it remains constant (whatever may be the initial concentration)
t1/2 = 2.5 hrs
For the reaction, 2NH3 → N2 + 3H2
(3/2)k1 = 3k2
= k3
1.5 k1 = 3k2 = k3
At low pressure the
reaction follows first order, therefore
Rate α [reactant]1
Rate α ( surface area )
At high pressure due to
the complete coverage of surface area, the reaction follows zero order.
Rate α[reactant]0
Therefore the rate is
independent of surface area.
Consider the following statements :
(i) increase in concentration of the reactant increases the rate
of a zero order reaction.
(ii) rate constant k is equal to collision frequency A if Ea
= 0
(iii) rate constant k is
equal to collision frequency A if Ea = °
(iv) a plot of ln(k) vs
T is a straight line
(v) a plot of ln (k) vs 1/T
is a straight line with a positive
slope.
In zero order reactions, increase in the concentration of reactant
does not alter the rate. So statement (i) is wrong.
k = A e – (Ea/RT)
if Ea = 0 so, statement
(ii) is correct, and statement (iii) is wrong
k = A e0
k = A
ln k = ln A – (Ea /R) (1/ T)
this equation is in the form of a straight line equatoion
y = c + m x
a plot of lnk vs 1/T is a straight line with negative slope
so statements (iv) and (v) are wrong.
(x+y) kJmol-1
(x+y) 103 Jmol-1