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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)

best_answer

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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