Let the equilibrium constant of Step I be 2 - 10-3 mol-1 L and the rate constants for the formation of A2X and A2 in Steps II and III be 3-0 - 10-2 mol-1 L min-1 and 1 - 103 mol-1 L min-1 all data at 25-C- Then- what is the overall rate constant mol-2 L2 min-1 for the consumption of B2-

The correct option is B 1.2 × 10^ 4 ...

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Standard XII

Chemistry

Solubility Product

Let the equil...

Question

Let the equilibrium constant of Step I be $2 \times 10^{- 3} m o l^{- 1} L$ and the rate constants for the formation of $A_{2} X$ and $A_{2}$ in Steps II and III be $3.0 \times 10^{- 2}$ $m o l^{- 1} L m i n^{- 1}$ and $1 \times 10^{3} m o l^{- 1} L m i n^{- 1}$ (all data at $25^{\circ} C$ ). Then, what is the overall rate constant $(m o l^{- 2} L^{2} m i n^{- 1})$ for the consumption of $B_{2}$ ?

6 \times 10^{- 5}

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1.2 \times 10^{- 4}

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3 \times 10^{- 5}

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1.5 \times 10^{- 5}

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Solution

The correct option is B $1.2 \times 10^{- 4}$

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Q. A reaction takes place in various steps. The rate constant for first, second, third and fifth steps are

k_{1}, k_{2}, k_{3}

and

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, if activation energy are

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respectively, the overall energy of activation (

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Q. Following data are given for the reaction between

A

and

B

$[A] / m o l L^{- 1}$	$[B] / m o l L^{- 1}$	Initial rate/ $m o l L^{- 1} s^{- 1}$ at $300 K$	$320 K$
$2.5 \times 10^{- 4}$	$3.0 \times 10^{- 5}$	$5.0 \times 10^{- 4}$	$2.0 \times 10^{- 3}$
$5.0 \times 10^{- 4}$	$6.0 \times 10^{- 5}$	$4.0 \times 10^{- 3}$	-
$1.0 \times 10^{- 3}$	$6.0 \times 10^{- 5}$	$1.6 \times 10^{- 2}$	-

The rate constant at

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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.

The ionization constant of $N H_{4}^{+}$ in water is $2.8 \times 10^{- 10}$ at $25^{\circ} C$ .The rate constant for the reaction of $N H_{4}^{+}$ and $O H^{-}$ to form $N H_{3}$ and $H_{2} O$ at $25^{\circ} C$ is $1.4 L m o l^{- 1} s^{- 1}$ ,the rate constant per proton transfer from water to $N H_{3}$ will be

Q. The reaction,

2 N_{2} O_{5} \to 4 N O_{2} + O_{2}

, was studied and the following data were collected:

Expt. No.	$[N_{2} O_{5}]$ $(m o l L^{- 1})$	Rate of disappearance of $N_{2} O_{5}$ $(m o l L^{- 1} {m i n}^{- 1})$
$1.$	$1.13 \times 10^{- 2}$	$34 \times 10^{- 5}$
$2.$	$0.84 \times 10^{- 2}$	$25 \times 10^{- 5}$
$3.$	$0.62 \times 10^{- 2}$	$18 \times 10^{- 5}$

Determine (i) order the reaction, (ii) the rate law, and (iii) rate constant for the reaction.

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The correct option is B 1.2×10−4

The correct option is B $1.2 \times 10^{- 4}$