The following results were obtained during kinetic studies of the reaction. 2 A + B → product Experiment A in   mol   L - 1 B   i n   m o l   L - 1 Initial rate of reaction in   mol   L - 1   min   - 1 I 0 . 10 0 . 20 6.93 × 10 - 3 II 0 . 10…

Question

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

2 A + B \to

product

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

The time (in minutes) required to consume half of

A

is

Aicharya · Accepted Answer

Let's assume x and y are the order of the reaction with respect to A and B , respectively. From equation 1 and 2 , 0 .1 0 .1 x 0 .2 0 .25 y = 6 .93 &#215; 10 - 3 6 .93 &#215; 10 - 3 &#8658; y = 0 From equation 1 and 3 and put the value y = 0 , 0 .1 0 .2 x 0 .2 0 .3 0 = 6 .93 &#215; 10 - 3 1 .386 &#215; 10 - 2 &#8756; y = 0 1 2 x = 1 2 &#8658; x = 1 The rate law will be R = K [ A ] 1 [ B ] 0 Hence, the reaction is of the first order. K = R A = 6 .93 &#215; 10 - 3 0 .1 = 6 .93 &#215; 10 - 2 Here t 1 2 for the first-order reaction is given by, t 1 2 = 0 .693 K = 0 .693 ( 6 .93 &#215; 10 - 2 ) =10

The following results were obtained during kinetic studies of the reaction. 2 A + B → product…

Solution

The following results were obtained during kinetic studies of the reaction. 2 A + B &#8594; product…

Solution

The following results were obtained during kinetic studies of the reaction. 2 A + B → product…