Difficult
Select the rate law that corresponds to the data shown for the following reaction $A + B \to C$
$Expt. \ No.$ $[A]$ $[B]$ $Initial \ Rate$ $1$ $0.012$ $0.035$ $0.10$ $2$ $0.024$ $0.070$ $0.80$ $3$ $0.024$ $0.035$ $0.10$ $4$ $0.012$ $0.070$ $0.80$
| $Expt. \ No.$ | $[A]$ | $[B]$ | $Initial \ Rate$ |
|---|---|---|---|
| $1$ | $0.012$ | $0.035$ | $0.10$ |
| $2$ | $0.024$ | $0.070$ | $0.80$ |
| $3$ | $0.024$ | $0.035$ | $0.10$ |
| $4$ | $0.012$ | $0.070$ | $0.80$ |
- ARate $= k[B]^3$
- BRate $= k[B]^4$
- CRate $= k[A][B]^3$
- DRate $= k[A]^2[B]^2$
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If the rate constant $k = 4.5 \times 10^{-7} \text{ L}^2 \text{ mol}^{-2} \text{ s}^{-1}$,then what is the order of the reaction?
MediumGUJCET 2026
View SolutionFor the reaction $X_{2(g)} + Y_{2(g)} \rightarrow 2XY_{(g)}$,the following data are observed:
$[X_{2}] \ (M)$ $[Y_{2}] \ (M)$ Rate of appearance of $XY \ (M \ sec^{-1})$ $0.1$ $0.1$ $5 \times 10^{-6}$ $0.2$ $0.1$ $10^{-5}$ $0.2$ $0.2$ $4 \times 10^{-5}$
Calculate the rate constant of the reaction (in $M^{1-n} \ sec^{-1}$),where $n$ is the order of the reaction.
| $[X_{2}] \ (M)$ | $[Y_{2}] \ (M)$ | Rate of appearance of $XY \ (M \ sec^{-1})$ |
| $0.1$ | $0.1$ | $5 \times 10^{-6}$ |
| $0.2$ | $0.1$ | $10^{-5}$ |
| $0.2$ | $0.2$ | $4 \times 10^{-5}$ |
Calculate the rate constant of the reaction (in $M^{1-n} \ sec^{-1}$),where $n$ is the order of the reaction.
Difficult
View SolutionThe reaction,$X + 2Y + Z \to N$ occurs by the following mechanism:
$(i)$ $X + Y \rightleftharpoons M$ (very rapid equilibrium)
$(ii)$ $M + Z \to P$ (slow)
$(iii)$ $P + Y \to N$ (very fast)
What is the rate law for this reaction?
$(i)$ $X + Y \rightleftharpoons M$ (very rapid equilibrium)
$(ii)$ $M + Z \to P$ (slow)
$(iii)$ $P + Y \to N$ (very fast)
What is the rate law for this reaction?
Difficult
View SolutionFor the reaction: $2 A + B \rightarrow A_2 B$,the rate $= k[A][B]^2$ with $k = 2.0 \times 10^{-6} \ mol^{-2} \ L^2 \ s^{-1}$. Calculate the initial rate of the reaction when $[A] = 0.1 \ mol \ L^{-1}$ and $[B] = 0.2 \ mol \ L^{-1}$. Calculate the rate of reaction after $[A]$ is reduced to $0.06 \ mol \ L^{-1}$.
Difficult
View SolutionThe reaction $2 \ NO + Br_2 \rightarrow 2 \ NOBr$ takes place through the mechanism given below:
$NO + Br_2 \Leftrightarrow NOBr_2$ (fast)
$NOBr_2 + NO \rightarrow 2 \ NOBr$ (slow)
The overall order of the reaction is $.....$.
$NO + Br_2 \Leftrightarrow NOBr_2$ (fast)
$NOBr_2 + NO \rightarrow 2 \ NOBr$ (slow)
The overall order of the reaction is $.....$.
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