Chemical Kinetics
Sponsor Area
A first-order reaction has a specific reaction rate of 10–2 s–1. How much time will it take for 20 g of the reactant to reduce to 5 g?
-
238.6 second
-
138.6 second
-
346.5 second
-
693.0 second
B.
138.6 second
For the reduction of 20 g of reactant to 5 g, two t1/2 is required.
Sponsor Area
Activation energy (Ea) and rate constants (k1 and k2) of chemical reaction at two different temperatures (T1 and T2) are related by
D.
According to Arrhenius equation, activation energy (Ea) and rate constants (k1 and k2) of chemical reaction at two different temperatures (T1 and T2) are related as,
![]()
Consider the reaction
N2 (g) + 3 H2 (g) → 2 NH3 (g)
The equality relationship between
is:
B.
For the reaction,
N2 (g) + 3 H2 (g) → 2 NH3 (g)

During the kinetic study of the reaction, 2A + B --> C+ D, following results were obtained
|
|
[A]/mol L- |
[B]/ mol L- |
Initial rate of formation of D/ mol L- min- |
|
I |
0.1 |
0.1 |
6.0 x 10-3 |
|
II |
0.3 |
0.2 |
7.2 x 10-2 |
|
III |
0.3 |
0.4 |
2.88 x 10-1 |
|
IV |
0.4 |
0.1 |
2.40 x10-2 |
based on the above data which one of the following is correct?
-
rate = k[A]2[B]
-
rate = k [A][B]
-
rate = k[A]2[B]2
-
rate = k[A][B]2
D.
rate = k[A][B]2
Let the order of reaction with respect to A is x and with respect to B is y. Thus,
rate = k[A]x[B]y
For the given cases,
(I) rate = k(0.1)x (0.1)y = 6.0 x 10-3
(II) rate = k (0.3)x (0.2)y = 7.2 x 10-2
(III) rate =k(0.3)x (0.40)y = 2.88 x 10-1
(IV) rate = k(0.4)x (0.1)y = 2.40 x 10-2
On dividing eq. (I) and (IV), we get

For an endothermic reaction, energy of activation is Ea and enthalpy of reaction is ΔH (both of these in kJ/mol). Minimum value of Ea will be
-
less than ΔH
-
equal to ΔH
-
more than ΔH
-
equal to zero
C.
more than ΔH
In endothermic reactions, the energy of reactants is less than that of the products. Potential energy diagram for endothermic reactions is,
Where Ea = activation energy of forwarding reaction
Ea' = activation energy of backwards reaction
ΔH = enthalpy of the reaction
From the above diagram,
Ea = Ea' + ΔH
Thus, Ea > ΔH
Sponsor Area
Mock Test Series
Mock Test Series



