The reaction of hydrogen peroxide with iodine, h2o2(aq)+i2(aq)⇌oh−(aq)+hio(aq) is first order in h2o2 and first order in i2. if the concentration of h2o2 was increased by half and the concentration of i2 was quadrupled, by what factor would the reaction rate increase?

So we can say that the rate of reaction increases by factor of 6.

Explanation:

The rate law for any given reaction  

A+B⇄C+D

Rate law for the above reaction is:

R=K[A]ᵃ[B]ᵇ

a and b are the order of reaction and it is an experimentally determined quantity.

K is the rate constant and it is constant for a given reaction

[A] and [B] are the concentrations of the reactants.

R is the rate of reaction

For the given reaction :

H₂O₂(aq.)+I₂(aq.)⇆OH⁻(aq.)+HIO(aq.)

Also it is given for the reaction that order with respect to H₂O₂ is 1 and order with respect to I₂ is also 1

The rate law can be written as :

R=K[H₂O₂]¹[I₂]¹

k=rate constant

When we increase the  concentration of H₂O₂ by half which meansthat new concentration of H₂O₂ will be= 3/2[H₂O₂].

When we increase the  concentration of I₂ by 4 which means that new concentration of I₂ will be= 4[I₂]

Putting the values of our new concentration in the rate law:

R=K3/2[H₂O₂]¹4[I₂]¹

R=6K[H₂O₂]¹[I₂]¹=New rate

So as we can see that the new rate of the reaction using the new concentration is 6 times the older rate of reaction.

So we can say that the rate of reaction increases by factor of 6.

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.