Question

On doubling the concentration of reactant the rate of reaction triples. Find the reaction order

Answer 1

Reaction order:

-$r_{A}$=k$C_{A} ^{n}$

3$(-r_{A} )$=k$(2C_{A} )^{n}$

$3(kC_{A} ^{n} )$=k$2^{n}$$C_{A} ^{n}$

3=$2^{n}$⇒ln3=n*ln2

n=ln/3/ln2

=1.5849

=1.6

Answer 2

The reaction order is 1.6

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

For reaction: $A\rightarrow products$

$Rate=k[A]^x$    (1)

k= rate constant

$[A]$ = initial conccentartion of A

Given :

$3\times Rate=k[2A]^x$   (2)

Dividing (2) by (1)

$\frac{3\times Rate}{Rate}=\frac{k[3A]^x}{k[A]^x}$

$3=\frac{k[2]^x[A]^x}{k[A]^x}$

$3=[2]^x$

$x=1.6$

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