units of rate constant of first and zero order reactions in terms of molarity M unit are respectively

  • Option 1)

    sec^{-1},Msec^{-1},

  • Option 2)

    sec^{-1},M

  • Option 3)

    Msec^{-1},sec^{-1},

  • Option 4)

    M,sec^{-1},

 

Answers (1)

As we learnt in

Zero Order Reaction -

Zero order reaction means that the rate of the reaction is proportional to zeroth power of the concentration of reactants

 

- wherein

R\rightarrow P

Rate=\frac{-d(R)}{dt}=K[R]^{0}=K

t_\frac {1}{2}= time required for half completion of reaction=\frac{a}{2k}

 

 

First Order Reaction -

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

- 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}

 

 

 for zero order reaction

-\frac{\Delta \left [ R \right ]}{\Delta t}=k\left [ R \right ]^{0}

unit of k is MSec^{-1}

for first order reaction

-\frac{\Delta \left [ R \right ]}{\Delta t}=k\left [ R \right ]^{1}

unit of  Sec^{-1}


Option 1)

sec^{-1},Msec^{-1},

Correct option

Option 2)

sec^{-1},M

Incorrect option

Option 3)

Msec^{-1},sec^{-1},

Incorrect option

Option 4)

M,sec^{-1},

Incorrect option

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