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For the reaction , the rate law is found to be Rate = . Which of the following mechanisms is consistent with this rate law?
(slow)
(fast)
(fast)
(slow)
(slow)
(fast)
(fast)
(slow)
The mechanism of a hypothetical reaction is given below:
(i) (fast)
(ii) (slow)
(iii) (fast)
What is the overall order of the reaction?
1
2
1.5
0.5
For the reaction , the following mechanism is proposed:
(i) (fast)
(ii) (slow)
What is the rate law for this reaction?
In a reaction mechanism with a fast equilibrium followed by a slow step, which step determines the overall rate law?
The fast equilibrium step
The slow step
Both steps equally
Neither step
The reaction follows the mechanism:
(i) (fast equilibrium, K)
(ii) (slow, k)
If the initial concentration of P is doubled while keeping the concentration of Q constant, how will the initial rate of formation of R change?
Doubles
Increases by a factor of
Quadruples
Remains unchanged
A reaction proceeds via a two-step mechanism: a fast reversible step followed by a slow step. If the overall order of the reaction is determined to be 1.5, what are possible orders of the reactants involved in the slow step?
1 and 1
2 and 0
1 and 0.5
0.5 and 0.5
The reaction obey I order with respect to and both
Which of the following mechanism is in consistent with the given fact?
Mechanism
Mechanism
Mechanism A
Mechanism B
Both Mechanism A and B
Neither Mechanism A nor B
Consider the reaction,
the rate equation for this reaction is ,
Which of these mechanisms is/are consistent with this rate equation?
$\begin{array}{l}
A.;;C{l_2} + {H_2}S\left( {aq} \right) o {H^ + }C{l^ - } + C{l^ + } + H{S^ - }\left( {slow} \right)\
;;;;;C{l^ + } + H{S^ - } o {H^ + }C{l^ - } + S\left( {fast} \right)\
B.;;{H_2}S \leftrightarrow {H^ + } + H{S^ - }\left( {fast;equilibrium} \right)\
;;;;C{l_2} + H{S^ - } o 2C{l^ - } + {H^ + }S\left( {slow} \right)
\end{array}$
(B) only
Both (A) and (B)
Neither (A) nor (B)
(A)Only