In a consecutive first order reaction,
(where k_{1} and k_{2} are the respective rate constants) species B has transient existence. Therefore,
The specific rate constant of decomposition of a compound is represented by:
The activation energy of decomposition for this compound at 300 K is
In a zeroorder reaction for every 10° rise of temperature, the rate is doubled. If the temperature is increased from 10°C to 100°C, the rate of the reaction will become
The fluorescence life time of a molecule in solution is 10 ns. If the fluorescence quantum yield is 0.1, the rate constant of fluorescence decay is
For the reaction, P + Q + R → S, experimental data for the measured initial rate is given below:
The order of the reaction with respect to P, Q and R respectively, is
Sucrose is converted to a mixture of glucose and fructose in a pseudo first order process under alkaline conditions. The reaction has a half˜life of 28.4 min. The time required for the reduction of a 8.0 mM sample of sucrose to 1.0 mM is:
The reaction, 2NO(g) + O_{2}(g) →2NO_{2}(g) proceeds via the following steps:
The rate of this reaction is equal to
1 g of ^{90}Sr gas converted to 0.953 g after 2 yr. The halfline of ^{90}Sr, and the amount of ^{90}Sr remaining after 5 yr are:
How many times will the rate of the elementary reaction 3X + Y → X2Y change if the concentration of the substance X is doubled and that of Y is halved?
Since it is an elementary reaction, its rate law r_{1} = k [A] ^{3}[B]
When the concentrations are changed the new rate will be r_{2} = k (2[A])^{3}([B]/2) = 4k[A]^{3}[B]
So, r_{2} = 4r_{1}.
A reaction proceeds through the formation of an intermediate B in a unimolecular reaction :The integrated rate law for this reaction is
In radical chain polymerization, the quantity given by the rate of monomer depletion, divided by the rate of propagating radical formation is called:
H_{2} and Br_{2} react to give HBr by the following steps:
The probable rate law for the above sequence is
The half–line of a order reaction varies with temperature according to:
k = ln2/t_{1/2} (i)
k = Ae^{(Ea/RT)} (ii)
Taking ln on both sides
lnk = ln((ln2)/ t_{1/2})
lnk = lnA + (E_{a}/R) x 1/T
Hence comparing the two equations, we get:
ln(t_{1/2}) α 1/T
Therefore A is the correct answer.
The half–life time for a reaction at initial concentrations of 0.1 and 0.4 mol^{–1} are 200 s and 50 s respectively. The order of the reaction is:
N_{2}O_{2}(g) and NO(g) react to form NO_{2} according to the stoichiometric equation
N_{2}O_{2}(g) + NO (g) →3NO_{2}(g) (R1)
At a given temperature and pressure. A possible mechanism for this overall reaction is:
Where NO_{3} is an unstable intermediate. What would be the rate expression of the disappearance of N_{2}O_{5} in terms of the concentrations of the stable species and the rate constants given above:
Formation of Z and X is theoretically expected to obey the following kinetic scheme.
An experimentalist wants to verify the above scheme, but can observed and measure the concentration of only X and Z. Is it possible that under certain, the measurements of [X] and [Z] as function of time would lead the experimenter to conclude that the kinetic scheme is as given below, and that the species Y is absent:
Following is the graph between (a – x)^{–1} and time t for a second order reaction,
Hence, rate at the start of the reaction is
Which of the following plots represent(s) the Arrhenius rate equation, k=Ae^{–Ea/RT} with and
Option D is correct as the exponential graph will be there in I and the negative slope will be there in II
The fraction of group condensed at time t in any stepwise condensation polymerization (overall second order) reaction is
For the following reaction,
According to Arrhenius equation (K= rate constant and T= Temperature):
The reaction of nitric oxide with ozone takes place according to the following stoichiometry.
The observed rate law for the reaction is found to be
Where α and β are constant. To explain the above rate low, the following reaction scheme has been proposed:
Determine an expression for β .
If the concept of halflife is generalized to quarter–life of a first order chemical reaction, it will e equal to:
The activation energy for the bimolecular reaction A+BC→AB+C is E_{0} in the gas phase. If the reaction is carried out in a confined volume of λ^{3} , the activation energy is expected to:
The minimum energy requirement that must be met for a chemical reaction to occur is called the activation energy. The activation energy depends on nature of the reacting species or reactants. It does not depends on volume that is it will remain constant irrespective of the volume.
So, (A) remain unchanged  is the correct option
In a reaction, A+B→Product, rate is doubled when the concentration of B is doubled and the rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled, rate law for the reaction can be written as:
When conc of B is doubled, the rate is doubled. So, Order wrt [B] is 1. The rate increases by a factor of 8 when conc of [A] and [B] are doubled. So Order wrt [A] is 2. Overall Rate is Rate = k[A]^{2}[B]
The expression for the equilibrium constant (K_{eq}) for the enzyme catalyzed reaction given below is:
For a reaction involving two steps given below:
Assume that the first step attains equilibrium rapidly. The rate of formation of P is proportional to
Consider the reaction H_{2} + C_{2} H_{2} → C_{2} H_{6} the molecular diameters of H_{2} and C_{2}H_{4} The preexponential factor in the rate constant calculated using collision theory in m^{3} (mol) ^{–1} s^{–1} is approximately (For this reaction at 300K, where the symbols have their usual meanings):
A reaction A → B , involves following mechanism:
The rate law of the reaction may be given as:
A decomposes as
Concentration of A, is equal to:




