The three experimental data for determine the differential rate of reaction $2 NO _{( g )}+ Cl _{2( g )} \rightarrow 2 NOCl_{( g )}$ at definate temperature. are given below.

$(a)$ Calculate order of reaction.

$(b)$ Calculate value of rate constant.

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$(a)$ Rate $=-\frac{d\left[\mathrm{Cl}_{2}\right]}{d t}=k[\mathrm{NO}]^{\frac{1}{2}}\left[\mathrm{Cl}_{2}\right]^{1}$ Order of reaction $=\frac{1}{2}+1=\frac{3}{2}=1.5$

$(b)$ $k=0.175\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^{-\frac{1}{2}} \mathrm{~s}^{-1}$

Similar Questions

The reaction between $A$ and $B$ is first order with respect to $A$ and zero order with respect to $B$. Fill in the blanks in the following table:

Experiment $[ A ] / mol\, ^{-1}$ $[ B ] / mol\, ^{-1}$ Initial rate $/$ $mol$ $L^{-1}$ $min$ $^{-1}$
$I$ $0.1$ $0.1$ $2.0 \times 10^{-2}$
$II$ - $0.2$ $4.0 \times 10^{-2}$
$III$ $0.4$ $0.4$ -
$IV$ - $0.2$ $2.0 \times 10^{-2}$

 

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

$(iv)$ $C _{2} H _{5} Cl ( g ) \rightarrow C _{2} H _{4}( g )+ HCl ( g ) \quad$ Rate $=k\left[ C _{2} H _{5} Cl \right]$

$2 NO +2 H _{2} \rightarrow N _{2}+2 H _{2} O$

The above reaction has been studied at $800^{\circ} C$. The related data are given in the table below.

Reaction serial number Initial pressure of $H _{2}$ / $kPa$ Initial Pressure of $NO /$ $kPa$ Initial rate $\left(\frac{- dp }{ dt }\right) /( kPa / s )$
$1$ $65.6$ $40.0$ $0.135$
$2$ $65.6$ $20.1$ $0.033$
$3$ $38.6$ $65.6$ $0.214$
$4$ $19.2$ $65.6$ $0.106$

 The order of the reaction with respect to $NO$ is $...........$ 

  • [JEE MAIN 2022]

For the reaction $2HI$ $\rightleftharpoons$ ${H_2} + {I_2}$, the rate of the reaction is proportional to ${[HI]^2}$. This means that the reaction is

During Kinetic study of reaction $2 A+B \rightarrow C+D$, the following results were obtained :

  $A[M]$ $B[M]$

initial rate of

formation of $D$

$i$ $0.1$ $0.1$ $6.0 \times 10^{-3}$
$ii$ $0.3$ $0.2$ $7.2 \times 10^{-2}$
$ii$ $0.3$ $0.4$ $2.88 \times 10^{-1}$
$iv$ $0.4$ $0.1$ $2.40 \times 10^{-2}$

Based on above data, overall order of the reaction is $\qquad$

  • [JEE MAIN 2024]