Question

The rate of a hypothetical reaction involving L and M is found to double when the concentration of L is doubled and to increase fourfold when the concentration of M is doubled. Write the rate law for this reaction.

Answer 1

Answer:
`Rate=k[L]^1[M]^2`

Step-by-step 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.

`L+M\rightarrow products`

`Rate=k[L]^x[M]^y` (1)

k= rate constant

x = order with respect to L

y = order with respect to M

n =( x+y)= Total order

a) If [L] is doubled, the reaction rate will increase by a factor of 2:

`2* Rate=k[2L]^x[M]^y` (2)

b) If [M] is doubled, the reaction rate will increase by a factor of 4:

`4* Rate=k[L]^x[2M]^y` (3)

Dividing 2 by 1:

`(2* Rate)/(Rate)=(k[2L]^x[M]^y)/(k[L]^x[M]^y)`

`2=2^x`

`x=1`

Dividing 3 by 1

`(4* Rate)/(Rate)=(k[L]^x[2M]^y)/(k[L]^x[M]^y)`

`4=2^y`

`2^2=2^y`

`y=2`

Thus rate law is:
`k[L]^1[M]^2`