# For a reaction $mA\rightarrow \: product$. Rate equation can be written as $rate = k[A]^{^{m}}$. If concn is changed from 0.1M to 0.15M then rate becomes double, what will be the value of m? Option 1) 0 Option 2) 1 Option 3) 1.7 Option 4) 2

As we learned in concept

Rate of Law = Dependence of Rate on Concentration -

The representation of rate of a reaction in terms of concentration of the reactants is known as Rate Law

or

The Rate Law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may/maynot be equal to stoichiometric of the reacting species in a balanced chemical equation

- wherein

Formula: $aA+bB\rightarrow cC+dD$

$Rate=\frac{dR}{dT}$

$=\alpha [A]^{x}.[B]^{y}$

$=k[A]^{x}.[B]^{y}$

K= rate constant

rate= $k[A]^{m}$ given

$r_1=k[A]^{m}= k(0.1)^{m}$

and  $r_2= k(0.5)^{m}$

Given that $r_2= 2r_1$

We have, $(0.15)^{m}= 2(0.1)^{m}$

$(1.5)^{m}=2$

$m\ln (1.5)= \ln(2)$

$m= \frac{0.693}{\ln(1.5)}=1.7$

Option 1)

0

Incorrect option

Option 2)

1

Incorrect option

Option 3)

1.7

Correct option

Option 4)

2

Incorrect option

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