The following results were obtained during the kinetic studies of the reaction- 2 A - B - P roducts- -A- -B- Initial rate of the reaction - Experiment in mol L -1- in mol L -1- in mol L -1- - min -1 -I - 0-10 0-20 -6-93 - 10-3II 0-100-256-93 - 10-3III 0-200-301-386 - 10-2The time in minutes required to consume half of A is-A- 5B- 100C- 1D- 10

The correct option is A 56.93×10^ 3 =K×(0.1)^x(0.2)^y 6.93×10^ 3 =K×(0.1)^x(0.25)^y So, y=0 and 1.386×10^ 2=K×(0.2)^x(0.30)^y x=1 so, r=K×(0.1)×(0.2)^0 6.93×10 ...

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Byju's Answer

Standard XII

Chemistry

Rate Constant

The following...

Question

The following results were obtained during the kinetic studies of the reaction: $2 A + B \to P r o d u c t s$

Experiment [A]
(in mol $L^{- 1}$ [B]
(in mol $L^{- 1}$ ) Initial rate of the reaction
(in mol $L^{- 1} m i n^{- 1}$ )

(I) 0.10 0.20 6.93x $10^{- 3}$

(II) 0.10 0.25 6.93x $10^{- 3}$

(III) 0.20 0.30 1.386x $10^{- 2}$

The time (in minutes) required to consume half of A is:

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Solution

The correct option is B 5
$6.93 \times 10^{- 3}$ = $K \times (0.1)^{x} (0.2)^{y}$
$6.93 \times 10^{- 3}$ = $K \times (0.1)^{x} (0.25)^{y}$
So, $y = 0$
and $1.386 \times 10^{- 2} = K \times (0.2)^{x} (0.30)^{y}$
$x = 1$
so, $r = K \times (0.1) \times (0.2)^{0}$
$6.93 \times 10^{- 3} = K \times 0.1 \times (0.2)^{0}$
K= $6.93 \times 10^{- 2}$
$t_{\frac{1}{2}}$ = $\frac{0.693}{2 K}$
= $\frac{10}{2}$ =5

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Similar questions

Q. The following results were obtained during kinetic studies of the reaction:

2 A + B \to

Products

Experment	[ $A$ ] (in mol $L^{- 1}$ )	[ $B$ ] (in mol $L^{- 1}$ )	Initial Rate of reaction (in mol $L^{- 1} m i n^{- 1}$ )
(I)	$0.10$	$0.20$	$6.93 \times 10^{- 3}$
(II)	$0.10$	$0.25$	$6.93 \times 10^{- 3}$
(III)	$0.20$	$0.30$	$1.386 \times 10^{- 2}$

The time (in minutes) required to consume half of

A

is:

Q. The following results were obtained during the kinetic studies of the reaction:

2 A + B \to P r o d u c t s

Experiment	[A] (in mol $L^{- 1}$	[B] (in mol $L^{- 1}$ )	Initial rate of the reaction (in mol $L^{- 1} m i n^{- 1}$ )
(I)	0.10	0.20	6.93x $10^{- 3}$
(II)	0.10	0.25	6.93x $10^{- 3}$
(III)	0.20	0.30	1.386x $10^{- 2}$

The time (in minutes) required to consume half of A is:

Q. During the kinetic study of the reaction,

2 A + B \to C + D

, following results were obtained:

Run	[A] ${mol L}^{- 1}$	[B] ${mol L}^{- 1}$	Initial rate of formation of D ${mol L}^{- 1} {min}^{- 1}$
I.	0.1	0.1	$6.0 \times 10^{- 3}$
II.	0.3	0.2	$7.2 \times 10^{- 2}$
III.	0.3	0.4	$2.88 \times 10^{- 1}$
IV.	0.4	0.1	$2.40 \times 10^{- 2}$

(Where,

[A]

and

[B]

are inital concentration of

A

and

B

respectively.)

Q. The following results have been obtained during the kinetic studies of the reaction:

2 A + B \to C + D

Experiment	$[A] / m o l L^{- 1}$	$[B] / m o l L^{- 1}$	Initial rate of formation of $D / m o l L^{- 1} m i n^{- 1}$
I	$0.1$	$0.1$	$6.0 \times 10^{- 3}$
II	$0.3$	$0.2$	$7.2 \times 10^{- 2}$
III	$0.3$	$0.4$	$2.88 \times 10^{- 1}$
IV	$0.4$	$0.1$	$2.40 \times 10^{- 2}$

Determine the rate law and the rate constant for the reaction.

The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D

Experiment	A/ mol L⁻¹	B/ mol L⁻¹	Initial rate of formation of D/mol L⁻¹ min⁻¹
I	0.1	0.1	6.0 × 10⁻³
II	0.3	0.2	7.2 × 10⁻²
III	0.3	0.4	2.88 × 10⁻¹
IV	0.4	0.1	2.40 × 10⁻²

Determine the rate law and the rate constant for the reaction.

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The correct option is B 56.93×10−3 =K×(0.1)x(0.2)y 6.93×10−3 =K×(0.1)x(0.25)y So, y=0 and 1.386×10−2=K×(0.2)x(0.30)y x=1 so, r=K×(0.1)×(0.2)0 6.93×10−3=K×0.1×(0.2)0 K=6.93×10−2 t12=0.6932K =102=5