For a reaction:
\( 2 \text{H}_2\text{O}_2 \xrightarrow{\text{I}} 2 \text{H}_2\text{O} + \text{O}_2 \)
The proposed mechanism is as given below:
(I) \( \text{H}_2\text{O}_2 \xrightarrow{\text{slow}} \text{H}_2\text{O} + \text{IO}^- \) (slow)
(II) \( \text{H}_2\text{O}_2 + \text{IO}^- \xrightarrow{\text{fast}} \text{H}_2\text{O} + \text{I}^+ + \text{O}_2 \) (fast)
(I) The rate-determining step is the first reaction, where hydrogen peroxide decomposes to form water and IO−. The rate law for the reaction will be dependent on the concentration of H2O2.
(II) In the second step, the intermediate IO− reacts with H2O2 in a fast step to produce water, iodine ions, and oxygen gas. This step does not affect the rate law, as it is not rate-determining.
(1) Rate law: The rate law is determined by the slow step, so the rate law is:
Rate = \( k[\text{H}_2\text{O}_2] \)
where \( k \) is the rate constant.
(2) Overall order and molecularity: The overall order of the reaction is 1, as the rate law depends on the concentration of only one reactant, H2O2. The molecularity of the reaction is 2, as the rate-determining step involves the collision of two molecules of H2O2.
Rate law for a reaction between $A$ and $B$ is given by $\mathrm{R}=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}$. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ is
For $\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftharpoons 2 \mathrm{AB}$ $\mathrm{E}_{\mathrm{a}}$ for forward and backward reaction are 180 and $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. If catalyst lowers $\mathrm{E}_{\mathrm{a}}$ for both reaction by $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Which of the following statement is correct?
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.