The predicted rate law, using the steady state approximation, for the reaction H₂O₂ + 2H⁺ + 2I^- → I₂ + 2H₂O following the possible mechanism is

rapid equilibrium

HI + H₂O₂  H₂O + HOI slow

HOI + I^-  I₂ + OH^- fast

OH^- + H⁺  H₂O fast is

Concept:-

• The rate of a chemical reaction is defined as the speed at which the reactants are converted into products. The rate of a reaction depends on the composition and the temperature of the reaction mixture.
• It is the amount of chemical change occurring with time.
• According to the steady state approximation, When a reaction proceeds steadily, there will be no overall accumulation of the intermediate, and there would be a stationary concentration of the same.
• When the concentration of the intermediate is very small, then according to the steady state approximation

​

• Let us consider the following consecutive reaction,

​

• If the concentration of the reactive intermediate 'B' has a small value that practically remains constant throughout the reaction, then we can apply steady state approximation for B as,

• Therefore for the consecutive reaction,

• Again, the Rate of the reaction is given by,

Explanation:-

For the reaction,

H2O2 + 2H+ + 2I- → I2 + 2H2O following the possible mechanism is

 rapid equilibrium

HI + H2O2  H2O + HOI (slow)

HOI + I-  I2 + OH- (fast)

OH- + H+  H2O (fast) is

• Now, the rate expression for HI is from the above equations;

So, 

or, 

or, ........(i)

• Now, From equation (i) for the slowest step we got

Rate = k₂[HI][H₂O₂]

= 

Conclusion:-

 Therefore the option 1 is correct.