Difficult
Fill in the blanks:
$1.$ The rate of a zero order reaction depends on the ........... concentration of the reactant.
$2.$ The molecularity of the slow step is equal to the ........... of the overall reaction.
$3.$ Rate $=$ ........ $[A]^x [B]^y$
$1.$ The rate of a zero order reaction depends on the ........... concentration of the reactant.
$2.$ The molecularity of the slow step is equal to the ........... of the overall reaction.
$3.$ Rate $=$ ........ $[A]^x [B]^y$
(A) $1.$ The rate of a zero order reaction is independent of the concentration of the reactant,meaning it depends on the $0^{th}$ power of the concentration.
$2.$ The molecularity of the slow step (rate-determining step) is equal to the order of the overall reaction.
$3.$ The rate law expression is given by $\text{Rate} = k[A]^x [B]^y$,where $k$ is the rate constant.
$2.$ The molecularity of the slow step (rate-determining step) is equal to the order of the overall reaction.
$3.$ The rate law expression is given by $\text{Rate} = k[A]^x [B]^y$,where $k$ is the rate constant.
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Similar Questions
Consider the following data for the given reaction $2 HI_{(g)} \rightarrow H_{2_{(g)}} + I_{2_{(g)}}$. The order of the reaction is:
Experiment $1$ $2$ $3$ $[HI] \ (mol \ L^{-1})$ $0.005$ $0.01$ $0.02$ Rate $(mol \ L^{-1} \ s^{-1})$ $7.5 \times 10^{-4}$ $3.0 \times 10^{-3}$ $1.2 \times 10^{-2}$
| Experiment | $1$ | $2$ | $3$ |
|---|---|---|---|
| $[HI] \ (mol \ L^{-1})$ | $0.005$ | $0.01$ | $0.02$ |
| Rate $(mol \ L^{-1} \ s^{-1})$ | $7.5 \times 10^{-4}$ | $3.0 \times 10^{-3}$ | $1.2 \times 10^{-2}$ |
DifficultJEE MAIN 2024
View SolutionFor the reaction $2A + B \rightarrow A_2B$,if the concentration of reactant $A$ is doubled and the concentration of reactant $B$ is halved,the rate of the reaction will:
Easy
View SolutionFor the reaction $2A + 2B \rightarrow 2C + D$,if $r = k[A]^2[B]^0$,then the rate of reaction is:
The rate constant for the reaction $A \longrightarrow B$ is $2 \times 10^{-4} \ L \ mol^{-1} \ min^{-1}$. The concentration of $A$ at which the rate of the reaction is $(1 / 12) \times 10^{-5} \ M \ sec^{-1}$ is :-
Medium
View SolutionThe correct difference between $1^{st}$-order and $2^{nd}$-order reaction is that:
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