In this case

k = x min^{−1} and [A_{0}]
= 0.01M = 1 × 10^{−2}M

t = 1 hour = 60 min

[A]= 1 × 10^{−2} ( e^{−60x})

k = A e ^{–(Ea/RT)}

ln k = ln A – (E_{a}/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

t_{1/2} = 0.693/k

t_{1/2} does not depend on the initial concentration and
it remains constant (whatever may be the initial concentration)

t_{1/2} = 2.5 hrs

For the reaction, 2NH_{3} → N_{2} + 3H_{2}

(3/2)k_{1} = 3k_{2}
= k_{3}

1.5 k_{1} = 3k_{2} = k_{3}

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.

A catalyst provides a new path to the reaction with low activation energy. i.e., it lowers the activation energy.

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 E_{a}
= 0

(iii) rate constant k is
equal to collision frequency A if E_{a} = °

(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 e^{0}

k = A

ln k = ln A – (E_{a} /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) 10^{3} Jmol^{-1}