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Second Order Reaction

The kinetic d...

Question

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 :

A

1 and 2

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B

2 and 1

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C

1 and 1

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D

2 and 2

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Solution

The correct option is D 1 and 2
When the concentration of B is kept constant and the concentration of A is doubled, the rate of the reaction is also doubled.
Hence, the reaction is first order in A.
When the concentration of A is kept constant and the concentration of B is doubled, the rate of the reaction becomes 4 times.
Hence, the reaction is second order in B.

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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 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:

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

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

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:

Q. 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	$[A] / m o l L^{- 1}$	$B] / m o l L^{- 1}$	Initial rate/ $m o l L^{- 1} m i n^{- 1}$
I	$0.1$	$0.1$	$2.0 \times 10^{- 2}$
II	-	$0.2$	$4.0 \times 10^{- 2}$
III	$0.4$	$0.4$	-
IV	-	$0.2$	$2.0 \times 10^{- 2}$

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