Chemistry · Elementary and complex reactions, order and molecularity of reactions, rate law, rate constant and its units
For the reaction given below, \rightarrow \mathbf{3} \mathbf{O}_{\mathbf{2}}(\ma
For the reaction given below, \( \mathbf{2} \mathbf{O}_{\mathbf{3}}(\mathbf{g}) \rightarrow \mathbf{3} \mathbf{O}_{\mathbf{2}}(\mathbf{g}) \) \( \operatorname{Step} \mid: \mathbf{O}_{3}(\mathbf{g}) \rightarrow \mathbf{O}_{2}(\mathbf{g})+\mathbf{O}(\mathbf{g}) \) Step \( 2: \mathrm{O}_{3}(\mathrm{g})+\mathrm{O} \rightarrow^{\text {slow }} 2 \mathrm{O}_{2}(\mathrm{g}) \) Statement I: The molecularity of the first step is 1 and of the second step is 2 Statement-II: \( O(g) \) is an intermediate and rate of reaction is \( \mathbf{K}\left[\mathbf{O}_{3}\right]^{2}\left[\mathbf{O}_{2}\right]^{-1} \) and order of reaction is 1
- A. Both statements are true Statement - - ll is correct explanation of Statement- -
- B. Both statements are true but Statement- -II is not correct explanation of Statement- -
- C. Statement - - lis true but Statement - II is false
- D. Statement-lis false but Statement- 11 is true
Step-by-step solution
Statement I is true: Step 1 involves one O3 molecule (molecularity 1), Step 2 involves O3 and O (molecularity 2). Statement II is also true: O is an intermediate, and the rate law derived using equilibrium approximation for step 1 (fast, reversible) gives rate = k[O3]^2[O2]^{-1} with overall order 1. However, Statement II does not explain why molecularity is 1 and 2; it provides additional information rather than a causal explanation.
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