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Published on: 29/08/2019
Chemical Kinetics
Download Tamil Nadu 12th Standard Chemistry question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Chemistry Test
1.
In pseudo-order reactions ______.
The actual order of reaction is different from that expected using rate law expression
The concentration of at least one reactant is taken in large excess.
The concentration of reactant taken in excess may be taken as constant
All of these
2.
For the reaction, 2N2O5 ⟶4 NO2+O2, select the correct statement.
Rate of formation of O2 is same as rate of formation of NO2
Rate of disappearance of N2O5 is two times the rate of formation of NO2.
Rate of formation of O2 is 0.5 times rate of disappearance of N2O5
Rate of formation of NO2 is equal to rate of disappearance of N2O5
3.
For a reaction, 2A + B ⟶ 3C, The rate of appearance of C at time 't' is 1.2 x 10-4 mol L-1s-1. Identify the rate of reaction.
4 x 10-5mol L-1s-1
4.5 x 10-1mol L-1s-1
3.6 x 10-4 mol L-1s-1
None of these
4.
The addition of a catalyst during a chemical reaction alters which of the following quantities?
Enthalpy
Activation energy
Entropy
Internal energy
5.
A zero order reaction X ⟶ Product, with an initial concentration 0.02M has a half life of 10 min. if one starts with concentration 0.04M, then the half life is
10 s
5 min
20 min
cannot be predicted using the given information
6.
For a first order reaction A ⟶ B the rate constant is x min−1. If the initial concentration of A is 0.01M, the concentration of A after one hour is given by the expression.
001. e−x
1 x 10-2(1-e-60x)
(1 x 10-2)e-60x
none of these
7.
CH3COCH3 + I2 \(\overset { H }{ \longrightarrow } \) ICH2 COCH3 + HI
8.
N2O5(g) ⟶ 2NO2(g) + \(\frac { 1 }{ 2 } \) O2(g)
9.
Arrhenius
10.
n = 2
1.
(d)
All of these
2.
(c)
Rate of formation of O2 is 0.5 times rate of disappearance of N2O5
3.
(a)
4 x 10-5mol L-1s-1
4.
A catalyst provides a new path to the reaction with low activation energy. i.e., it lowers the activation energy.
5.
for n ≠ 1 t1/2 = \(\frac{2^{n-1} -1}{(n- 1) k[A_{0}]^{-1}}\)
for n = 0; t1/2 = \(\frac{1}{2 k[A_{0}]^{-1}}\)
t1/2 = \(\frac{[A_{0}]}{2 k}\)
t1/2 α [A0] ...(1)
Given [A0] = 0.002 M; t1/2 = 10 min
[A0] = 0.04M; t1/2 = ?
Substitute in (1)
10 min α 0.02 M...(2)
t1/2 α 0.04M ....(3)
(3)(2)
⇒ t1/2 / 10 min
= 0.04 M/0.02 M
t1/2 = 2 x 10 min = 20 min
6.
\(k=\frac { 2.303 }{ t } log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(k=\frac { 1 }{ t } ln \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(e^{kt}=\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] }\)
[A] = [A0] ekt
In this case
k = x min-1 and [A0] = 0.01 M = 1 x 10-2M
t = 1 hour = 60 min
[A] = (1 x 10-2)e-60x
7.
0
8.
1
9.
\(k={ Ae }^{ -\left( \frac { Ea }{ RT } \right) }\)
10.
II Order
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