The kinetic datas for the reaction- 2A-B2- 2AB are as given below- -A- mol L-1 -B2-mol L-1Rate mol L-1 min-10-51-0 2-5- 10-31-01-0 5-0- 10-30-52-0 1- 10-2Hence the order of reaction with respect to A and B2 are- respectively-

The correct option is B 1 and 2 2.5× 10^ 3=k[0.5]^a [1.0]^β .....(1) 5× 10^ 3=k[1.0]^a [1.0]^β ......(2) 1× 10^ 2=k[0.5]^a [2.0]^β ....(3) Dividi ...

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Standard XII

Physics

Second Law of Motion

The kinetic d...

Question

The kinetic datas for the reaction: $2 A + B_{2} \to 2 A B$ are as given below:

$[A] m o l L^{- 1}$ $[B_{2}] m o l L^{- 1}$ Rate mol $L^{- 1} m i n^{- 1}$
0.5 1.0 $2.5 \times 10^{- 3}$
1.0 1.0 $5.0 \times 10^{- 3}$
0.5 2.0 $1 \times 10^{- 2}$

Hence the order of reaction with respect to A and $B_{2}$ are, respectively:

1 and 2

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2 and 1

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Solution

The correct option is B 1 and 2
$2.5 \times 10^{- 3} = k [0.5]^{a} [1.0]^{β} . . . . . (1)$
$5 \times 10^{- 3} = k [1.0]^{a} [1.0]^{β} . . . . . . (2)$
$1 \times 10^{- 2} = k [0.5]^{a} [2.0]^{β} . . . . (3)$
Dividing equation (1) and (2),
$\frac{1}{2} = [\frac{1}{2}]^{a}$
Hence $a = 1$ .
Dividing equation (1) and (3),
$\frac{2.5 \times 10^{- 3}}{1 \times 10^{- 2}} = (\frac{1.0}{2.0})^{β}$
$\frac{1}{4} = (\frac{1}{2})^{β}$

$\Rightarrow β = 2.$

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Similar questions

Q. The kinetic datas for the reaction :

2 A + B_{2} \to 2 A B

are as given below:

$[A]$ $m o l L^{- 1}$	$[B_{2}]$ $m o l L^{- 1}$	Rate $m o l L^{- 2} m i n^{- 1}$
$2$	$1.0$	$2.5 \times 10^{- 3}$
$1.0$	$1.0$	$2.5 \times 10^{- 3}$
$2$	$2.0$	$1 \times 10^{- 2}$

Hence the order of reaction w.r.t

A

and

B_{2}

are _____ respectively.

The kinetic datas for the reaction :

2 A + B_{2} \to 2 A B

arre as given below:

\begin{matrix} [A] (m o l L^{- 1}) & [B_{2}] (m o l L^{- 1}) & R a t e (m o l L^{- 1} m i n^{- 1}) 0.5 & 1.0 & 2.5 \times 10^{- 3} 1.0 & 1.0 & 5.0 \times 10^{- 3} 0.5 & 2.0 & 1 \times 10^{- 2} \end{matrix}

Hence the order of reaction with respect to

A

and

B_{2}

are respectively :

Q. The experimental data for the reaction,

2 A + B_{2} ⟶ 2 A B

, is as follows:

Expt. No.	$[A]$ $(m o l L^{- 1})$	$[B_{2}]$ $(m o l L^{- 1})$	Rate $(m o l L^{- 1} s^{- 1})$
$1.$	$0.50$	$0.50$	$1.6 \times 10^{- 4}$
$2.$	$0.50$	$1.00$	$3.2 \times 10^{- 4}$
$3.$	$1.00$	$1.00$	$3.2 \times 10^{- 4}$

Write the most probable equation for the rate of reaction giving reason for your answer.

Q. The experimental data for the reaction

2 A + B_{2} ⟶ 2 A B

, is:

Expt. No.	$[A]$	$[B_{2}]$	Rate $(m o l L^{- 1} s^{- 1})$
$1.$	$0.50$	$0.50$	$1.6 \times 10^{- 4}$
$2.$	$0.50$	$1.00$	$3.2 \times 10^{- 4}$
$3.$	$1.0$	$1.00$	$3.2 \times 10^{- 4}$

The rate equation for the above data is:

Expt. No.	$[A]$ $(m o l L^{- 1})$	$[B]$ $(m o l L^{- 1})$	Initial rate $(m o l L^{- 1} s^{- 1})$
Expt. No.	$[A]$ $(m o l L^{- 1})$	$[B]$ $(m o l L^{- 1})$	300K	320K
(1)	$2.5 \times 10^{- 4}$	$3.0 \times 10^{- 5}$	$5.0 \times 10^{- 4}$	$2.0 \times 10^{- 3}$
(2)	$5.0 \times 10^{- 4}$	$6.0 \times 10^{- 5}$	$4.0 \times 10^{- 3}$	$. . .$
(3)	$1.0 \times 10^{- 3}$	$6.0 \times 10^{- 5}$	$1.6 \times 10^{- 2}$	$. . .$

From the data given above for the reaction between

A

and

B

, calculate the following:
(i) The order of the reaction with respect to

A

and with respect to

B

,
(ii) The rate constant at

300

K

(iii) The energy of activation and
(iv) The pre-exponential factor.

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$[A] m o l L^{- 1}$	$[B_{2}] m o l L^{- 1}$	Rate mol $L^{- 1} m i n^{- 1}$
0.5	1.0	$2.5 \times 10^{- 3}$
1.0	1.0	$5.0 \times 10^{- 3}$
0.5	2.0	$1 \times 10^{- 2}$

The kinetic datas for the reaction: 2A+B2→2AB are as given below:[A]molL−1[B2]molL−1Rate mol L−1min−10.51.02.5×10−31.01.05.0×10−30.52.01×10−2Hence the order of reaction with respect to A and B2 are, respectively:

The correct option is B 1 and 22.5×10−3=k[0.5]a[1.0]β.....(1)5×10−3=k[1.0]a[1.0]β......(2)1×10−2=k[0.5]a[2.0]β....(3)Dividing equation (1) and (2),12=[12]aHence a=1.Dividing equation (1) and (3),2.5×10−31×10−2=(1.02.0)β14=(12)β ⇒β=2.