Answer:
Consider the reaction of hydrogen peroxide with iodine: H2O2(aq)+I2(aq)⇌OH−(aq)+HIO(aq) The reaction is first order in I2 and second order overall.
What is the rate law?
If the concentration of H2O2 is increased by half and the concentration of I2 is quadrupled, by what factor does the reaction rate increase?
Rate law = k[H₂O₂] [I₂]
The new rate is six times that of the old rate
Explanation:
Rate law = k[H₂O₂] [I₂]
lets input
[H₂O₂] = x
[I₂] = y
Rate = kxy
[H₂O₂] = x + x/2
= 3x/2
[I₂] = 4y
New rate = [tex]k * \frac{3x}{2} *4y[/tex]
New rate = 6
The new rate is six times that of the old rate
