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?

When H2O2 & i2 is a first order and the rate can be as the following:
rate = K [H2O2]^1 [i2]^1
and by assuming that [H2O2] & [i2] = 1 so by substitution 
the rate = K (1) (1) = 1 
and when H2O2 increase by half it becomes 1.5 and when i2 increase by quadruple it becomes 4.
by substitution:
the rate = K (1.5)(4) = 6 
So the rate will increase by factor 6

Idon't know hope someone you

Ibelieve the answer 4 moles