Practice free Chemical Kinetics (Chemistry) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
The rate constant of a reaction has same units as the rate of reaction. The reaction is of ...
$ zero order K = { rate \over [R]^n } K = rate , when \;n = 0 $
The rate constant of reaction is $ 3 \times 10^{-3} bar^{-1} sec^{-1} $ . The order of reaction is ...
$ 2 K = { rate \over [R]^n } = { bar /s \over bar^n } when n = 2 , k = bar ^ {-1} S^{-1} $
Which of the following statements is incorrect about the molecularity of a reaction ?
There is no difference between order and molecularity of a reaction.
For a single step reaction $ A + 2B ---> Products $ , the molecularity is
3
Which of the following statement is false ?
Both order and molecularity of a reaction are always the same.
If ‘a’ is the initial concentration of the reactant, the time taken for completion of the reaction, it if is of zero order, will be
a/k For Zero order reaction $ t = { 1 \over k} { [A]_o - [A] } But [A]_o = a $ And when reaction complete $ [A] = 0 \therefore t = { a \over k } $
The reaction $ 2O_3 ---> 3O_2 $ proceeds in two steps as follows. (i) $ O3 --> O_2 + O (fast) $ (ii) $ O + O_3 --> 2O_2 (slow) $ The rate law expression should be...
$ r = k [O_3]^2 [O_2] ^ {-1} From slowest step r = k [ O_3 ] [O] $ $ From eq (i) keq = [O_2] [O] / [O_3 ] $ $ \therefore [O] = Keq [O_3] / [O_2] _2 $ $ \therefore r = k [O_3] Keq [O_2] = K^1 [O_3]^2 [O_2]^{-1} $
For the reaction Zero order
$ t _ { 1/2} \alpha C_o $
For reaction first order
$ t_ { 1/2} = { 0.693 \over k } $
Which of the following represents the expression for ¾th life of a first order reaction
$ {2.303 \over k } log 4 $ $ t_{3/4} = { 2.303 \over k} log { ao \over ao -ao \times {3 \over 4} } = { 2.303 \over k } log { ao \over { ao /4 } } $
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