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: $Rate=k[L]^1[M]^2$

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:

$2\times Rate=k[2L]^x[M]^y$   (2)

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

$4\times Rate=k[L]^x[2M]^y$   (3)

Dividing 2 by 1:

$\frac{2\times Rate}{Rate}=\frac{k[2L]^x[M]^y}{k[L]^x[M]^y}$

$2=2^x$

$x=1$

Dividing 3 by 1

$\frac{4\times Rate}{Rate}=\frac{k[L]^x[2M]^y}{k[L]^x[M]^y}$

$4=2^y$

$2^2=2^y$

$y=2$

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