The initial rates of decrease of I_2 in aceto | vedclass

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(C) Let the rate law be $r = k[I_2]^x [H^{+}]^y [CH_3COCH_3]^z$.Comparing experiments $2$ and $3$: When $[H^{+}]$ and $[CH_3COCH_3]$ are constant,doub

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$0, 1, 1, 2$

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(C) Let the rate law be $r = k[I_2]^x [H^{+}]^y [CH_3COCH_3]^z$.Comparing experiments $2$ and $3$: When $[H^{+}]$ and $[CH_3COCH_3]$ are constant,doubling $[I_2]$ ($0.01$ to $0.02$) does not change the rate $(0.192)$. Thus,$2^x = 1 \Rightarrow x = 0$.Comparing experiments $1$ and $2$: When $[I_2]$ and $[CH_3COCH_3]$ are constant,doubling $[H^{+}]$ ($0.1$ to $0.2$) doubles the rate ($0.096$ to $0.192$). Thus,$2^y = 2 \Rightarrow y = 1$.Comparing experiments $2$ and $4$: When $[I_2]$ and $[H^{+}]$ are constant,doubling $[CH_3COCH_3]$ ($0.1$ to $0.2$) doubles the rate ($0.192$ to $0.384$). Thus,$2^z = 2 \Rightarrow z = 1$.Total order $= x + y + z = 0 + 1 + 1 = 2$.The orders are $0, 1, 1, 2$.

Similar Questions

For the reaction,$2N_2O_5 \to 4NO_2 + O_2$,the rate and rate constant are $1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$ and $3.4 \times 10^{-5} \ s^{-1}$ respectively. The concentration of $N_2O_5$ in $mol \ L^{-1}$ will be:

What is the order of reaction for the rate law $r = k [A]^{\frac{3}{2}} [B]^2$ ?

The rate for the reaction $2 A + B \rightarrow \text{product}$ is $6 \times 10^{-4} \ mol \ dm^{-3} \ s^{-1}$. Calculate the rate constant if the reaction is first order in $A$ and zeroth order in $B$,given $[A] = [B] = 0.3 \ M$.

$A$ reaction $2NO + 2H_2 \longrightarrow N_2 + 2H_2O$ has the following mechanism:
Step-$I$: $2NO \longrightarrow N_2O_2$
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Which of the following substances is a reaction intermediate?

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Experiment	Initial $[I_2]$ $(mol \ L^{-1})$	Initial $[H^{+}]$ $(mol \ L^{-1})$	Initial $[CH_3COCH_3]$ $(mol \ L^{-1})$	Initial rate $(mol \ L^{-1} \ s^{-1})$
$1$	$0.01$	$0.1$	$0.1$	$0.096$
$2$	$0.01$	$0.2$	$0.1$	$0.192$
$3$	$0.02$	$0.2$	$0.1$	$0.192$
$4$	$0.01$	$0.2$	$0.2$	$0.384$