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Ozone decomposes according to the equation
2O_{3}(g) → 3O_{2}(g) Mechanism of the reaction is
Step I : O_{3}(g) O_{2}(g) + O(g) (fast)
Step II : O_{3}(g) + O(g) → 2O_{2}(g) (slow)
Which of the following is correct ?
Slowest step is rate determing step
rate = K[O_{3}][O] .........(1)
Since, [O] is not in the original reaction hence it has to be eliminated.
From first step
Putting the value of [O] in the eq. (1)
Consider the following case of competing 1st order reactions After the start of the reaction at t = 0, with only P, concentration of Q is equal to R at all times. The time in which all the three
concentration will be equal is given by
Since [Q] = [R] all the times,
Taking ln of both side
–2K_{1}t = – ln3
As K_{1} = K_{2}
A substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as
The % distribution of B and C is ................ and .................. respectively.
K_{1} = 1.26 × 10^{–4} sec^{–1}
K_{2} = 3.8 × 10^{–5} sec^{–1}
Energy of activation can never be negative Option (D) is incorrect.
The basic theory behind Arrhenious equation is that
Only ans will be given
Which of the following statements are correct about half life period ?
(A) For zero order reaction
⇒ t_{1/2} α a_{0} (initial concentration)
(B) For first order reaction
Average life = 1.44 × half life
(C) For II^{nd} order reaction
t_{99.9} = 100 min
For a reaction : 2A + 2B → products, the rate law expression is r = k[A]^{2} [B]. Which of the following is/are correct?
Rate = K[A]^{2}[B]
⇒ rate w.r.t B = 1
Overall order of the reaction = 2 + 1 = 3
for a complex reaction the rate determing step is determined experimentally.
Which of the following statements are correct ?
In SN^{1} reaction, formation of carbocation takes place which is planar.
⇒ equimolar mixture of (I) and (II) will be formed in case of SN^{1}.
In SN^{2} reaction Nucleophile will attack from the opposite to leaving group
⇒ Product (II) will be formed in case of SN^{2} reaction.
Rate constant k varies with temperature by equation log_{10}k(min^{1}) = 5 – ^{2000}/_{T}. We can conclude
log A = 5 ⇒ A = 10^{5}
E_{a} = 2000 × 2.303 × 8.314 cal
E_{a} = 9.212 Kcal
A reaction is catalysed by H^{+} ion. In presence of HA, rate constant is 2 × 10^{3} min^{1}and in presence of HB rate constant is 1 × 10^{3} min^{1}. HA and HB being strong acids, we may conclude that
Rate = K[H^{+}][A]
Rate = K’[A]
If K’ has greater value for HA
⇒ [H^{+}] has to be large
⇒ for that acid has be stronger
⇒ [HA] is strongers that HB
Also Relative strength = 2
The elements which are good catalysts and have the ability to change their oxidation number are
Statement1 : The time of completion of reactions of type A → product (order <1) may be determined.
Statement2 : Reactions with order ³ 1 are either too slow or too fast and hence the time of completion can not be determined.
eg. Zero order reaction completion time can be determind for order > 1 time can be determined
Statement1 : Temperature coefficient of an one step reaction may be negative.
Statement2 : The rate of reaction having negative order with respect to a reactant decreases with the increase in concentration of the reactant.
Ea > 0
If order
Statement1 : The overall rate of a reversible reaction may decrease with the increase in temperature.
Statement2 : When the activation energy of forward reaction is less than that of backward reaction, then the increase in the rate of backward reaction is more than that of forward reaction on increasing the temperature.
Statement1 : In a reversible endothermic reaction, E_{act} of forward reaction is higher than that of backward reaction
Statement2 : The threshold energy of forward reaction is more than that of backward reaction
∴ ΔH > 0
A hypothetical elementary reaction where Initially only 2 moles of A are present.
The total number of moles of A, B & C at the end of 50% reaction are
2 – x – y = 1
⇒ x + y = 1 ............(1)
⇒ [B] = [C]
⇒ 2x = y .............(2)
Solving (1) and (2)
Total moles after 50 % of the reaction
= 2 – x – y + 2x + y
= 2 + x = 7/3
A hypothetical elementary reaction where Initially only 2 moles of A are present.
Number of moles of B are
No. of Moles of B = 2x
= 2/3
= 0.666
The rate of a reaction increases significantly with increase in temperature. Generally, rates of reaction are doubled for every 10º rise in temperature. Temperature coefficient gives us an idea about the change in rate of a reaction for every 10º change in temperature.
Temperature coefficient
Arrhenious gave an equation which describes rate constant k as a function of temperature is k = A^{ –Ea/RT}
where k is a rate constant
A is frequency factor or pre exponential factor
E_{a} is activation energy
T is temperature in kelvin and
R is universal gas constant
Equation when expressed in logarithmic form becomes
Q.
For a reaction E_{a} = 0 and k = 3.2 × 10^{5} s^{1} at 325 K. The value of k at 335 K would be
When Ea = 0, k = 3.2 ×10^{5} & it does not depend on T
The rate of a reaction increases significantly with increase in temperature. Generally, rates of reaction are doubled for every 10º rise in temperature. Temperature coefficient gives us an idea about the change in rate of a reaction for every 10º change in temperature.
Temperature coefficient
Arrhenious gave an equation which describes rate constant k as a function of temperature is k = A^{ –Ea/RT}
where k is a rate constant
A is frequency factor or pre exponential factor
E_{a} is activation energy
T is temperature in kelvin and
R is universal gas constant
Equation when expressed in logarithmic form becomes
Q.
For which of the following reactions k_{310}/k_{300} would be maximum ?
When E_{a} = is max,
The rate of a reaction increases significantly with increase in temperature. Generally, rates of reaction are doubled for every 10º rise in temperature. Temperature coefficient gives us an idea about the change in rate of a reaction for every 10º change in temperature.
Temperature coefficient
Arrhenious gave an equation which describes rate constant k as a function of temperature is k = A^{ –Ea/RT}
where k is a rate constant
A is frequency factor or pre exponential factor
E_{a} is activation energy
T is temperature in kelvin and
R is universal gas constant
Equation when expressed in logarithmic form becomes
Q.
Activation energies of two reaction are E_{a} and E_{a} with E_{a} > E_{a}. If the temperature of the reacting systems is increased from T_{1} to T_{2} (k` are rate constants at higher temperature).
The rate of a reaction increases significantly with increase in temperature. Generally, rates of reaction are doubled for every 10º rise in temperature. Temperature coefficient gives us an idea about the change in rate of a reaction for every 10º change in temperature.
Temperature coefficient
Arrhenious gave an equation which describes rate constant k as a function of temperature is k = A^{ –Ea/RT}
where k is a rate constant
A is frequency factor or pre exponential factor
E_{a} is activation energy
T is temperature in kelvin and
R is universal gas constant
Equation when expressed in logarithmic form becomes
Q.
For the reactions, following data is given
Temperature at which k_{1} = k_{2} is
The activation energy of a reaction at a given temperature is found to be 2.303 RT J mol^{–1}. The ratio of rate constant to the Arrhenius factor is
An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5° while after completion of the reaction it was –20°. If optical rotation per mole of A, B & C are 60°, 40° & –80°. Calculate half life of the reaction.
Optical rotation reaction no. of moles
At, t = 20 min
60 (a – x) + 40 x – 80 x = 5
60a – 100 x = 5
30 – 100 x = 5
No. of moles of A is becoming half after 20 min
∴ t_{1/2} = 20 min
Consider the following first order decomposition process:
Here, "t" corresponds to the time at which (^{1}/_{6})^{th} of reactant is decomposed. The value of "n" is
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