Question

For a reaction a+b gives product the rate of reaction was doubled when concentration of a was doubled. When concentration of a and b both was doubled, the rate was double the order of reaction wrt a and b are

Answer 1

if conc of a is doubled then rate of rxn is doubled it means order of a is 1 and when both double then rate is also double it means overall order of rxn is 1

• a+b =1
• and a=1 then b= 0 which is required

Answer 2

The order of reaction with respect to a and b are 1 and 0 respectively.

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.

$a+b\rightarrow products$

$Rate=k[a]^x[b]^y$    (1)

k= rate constant

x = order with respect to a

y = order with respect to b

n = x+y = Total order

a) From trial 1: $2\times Rate=k[2a]^x[b]^y$    (2)

From trial 2: $2\times Rate=k[2a]^x[2b]^y$    (3)

Dividing 3 by 2 :$\frac{2\times Rate}{2\times Rate}=\frac{k[2a]^x[2b]^y}{k[2a]^x[b]^y}$

$1=2^y,2^0=2^y$ therefore y=0.

b) $Rate=k[a]^x[b]^y$    (1)

From trial 2: $2\times Rate=k[2a]^x[b]^y$    (3)

Dividing  3 by 1 :$\frac{2\times Rate}{Rate}=\frac{k[2a]^x[b]^y}{k[a]^x[b]^y}$

$2=2^x,2^1=2^x$, x=1

Thus rate law is $Rate=k[a]^1[b]^0$

Thus order with respect to a is 1 , order with respect to b is 0 and total order is 1+0=1

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