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The following results were obtained during kinetic studies of the reaction

2A+B→Product:

Experiment
[A]

(in mole L^-1

[B]

(in mole L^-1

Initial rate of reaction

(in mole L^-1min^-1)

I 0.10 0.20
6.93 X 10^-3

II 0.10 0.25
6.93 X 10^-3

III 0.20 0.30 1.386 X 10^-2

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

Option 1)
5
Option 2)
10
Option 3)
1
Option 4)
100

Answers (1)

best_answer

Chemical Kinetics -

The branch of chemistry which predicts the rate and mechanism of a process

-

First Order Reaction -

The rate of the reaction is proportional to the first power of the concentration of the reactant.

- wherein

Formula:

R $\rightarrow$ P

a 0

a-x x

$rate[r]=K[R]^{1}$

$\frac{-d(a-x)}{dt}=K(a-x)$

$\frac{-dx}{dt}=K(a-x)$ [differentiate rate law]

$ln \:[\frac{a}{a-x}]=kt \:(Integrated rate law)$

Unit of $k=sec^{-1}$

$t_\frac{1}{2}=\frac{0.693}{k}$

As we know that:

rate= $KA^{x}B^{y}$

$6.93\times 10^{-3}=K\left ( 0.1 \right )^{x}\left ( 0.2 \right )^{y}---1$

$6.93\times 10^{-3}=K\left ( 0.1 \right )^{x}\left ( 0.25 \right )^{y}---2$

$so, y=0\, after \, solving \, 1\, and\, 2$

THen $1.386\times 10^{-2}=K\left ( 0.2 \right )^{x}\left ( 0.3 \right )^{y}---3$

1/3 then

$\frac{1}{2}=\left ( \frac{0.1}{0.2} \right )^{x}=> x=1$

rate= $K\left ( 0.1 \right )^{1}\left ( 0.2 \right )^{0}$

$6.93\times 10^{-3}=K\left ( 0.1 \right )$

$K=6.93\times 10^{-2}$

$tr_{2}=\frac{ln_{2}}{K_{2}}=\frac{0.693}{0.693\times 10^{-1}\times 2}=5minuts$

asking half of A not 2A

Option 1)
5
Option 2)

10

Option 3)

1

Option 4)

100

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