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

\(H^{+}+I^{-} \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} H I\) rapid equilibrium

HI + H₂O₂ \(\xrightarrow{K_2}\) H₂O + HOI slow

HOI + I^- \(\xrightarrow{K_3}\) I₂ + OH^- fast

OH^- + H⁺ \(\xrightarrow{K_4}\) 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

​\(\frac{d\left [ Concentartion \right ]}{dt} = 0\)

• Let us consider the following consecutive reaction,

​\(A\overset{K_1}{\rightarrow}B\overset{K_2}{\rightarrow}C\)

• 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,

\(\frac{d\left [B \right ]}{dt} = 0\)

• Therefore for the consecutive reaction,

\(A\overset{K_1}{\rightarrow}B\overset{K_2}{\rightarrow}C\)

\(\frac{d\left [ B \right ]}{dt}=K_1\left [ A_1 \right ]-K_2\left [ B \right ]=0\)

\(\left [ B \right ]=\frac{K_1}{K_2}\left [ B \right ]\)

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

\(\frac{d\left [ C \right ]}{dt}=K_2\left [ B \right ]\)

\(=K_2\times \frac{K_1}{K_2}\left [ A \right ]\)

\(=K_1 [A]\)

Explanation:-

For the reaction,

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

\(H^{+}+I^{-} \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} H I\) rapid equilibrium

HI + H2O2 \(\xrightarrow{K_2}\) H2O + HOI (slow)

HOI + I- \(\xrightarrow{K_3}\) I2 + OH- (fast)

OH- + H+ \(\xrightarrow{K_4}\) H2O (fast) is

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

\(\frac{d[HI]}{dt}=0\)

So, \(\frac{d[HI]}{dt} =K_1[H]^{+}[I]-K_{-1}[HI]-K_2[H_2O_2][HI]\)

or, \(K_1[H]^{+}[I]-K_{-1}[HI]-K_2[H_2O_2][HI]=0\)

or, \([HI]=\frac{k_1[H^+][I^-]}{k_{-1}+k_2[H_2O_2]}\)........(i)

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

Rate = k₂[HI][H₂O₂]

= \(\frac{k_1k_2[H^+][I^-][H_2O_2]}{k_{-1}+k_2[H_2O_2]}\)

Conclusion:-

 Therefore the option 1 is correct.