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)
A admin

 

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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