For the chemical reaction
`I^(ɵ) + OCl^(ɵ) rarr Cl^(ɵ) + OI^(ɵ)`
Rate `= (k[OCl^(ɵ)] [I^(ɵ)])/([overset(ɵ)(OH)]`
a. What is the order and moleculartiy of the reaction ?
b. In the above reaction, what are the molecularity and order with respect to `[overset(ɵ)(OH)]`

Options

A. `1,2`
B. `2,1`
C. `2,2`
D. `3,2`

Answer

To solve the problem, we need to analyze the given chemical reaction and its rate law.

### Given:
**Chemical Reaction:**
\[ I^{\circ} + OCl^{\circ} \rightarrow Cl^{\circ} + OI^{\circ} \]

**Rate Law:**
\[ \text{Rate} = \frac{k [OCl^{\circ}] [I^{\circ}]}{[OH^{\circ}]} \]

### Part A: Order and Molecularity of the Reaction

1. **Identify the Order of the Reaction:**
- The order of a reaction is determined by the sum of the powers of the concentration terms in the rate law.
- From the rate law, we see:
- The concentration of \( OCl^{\circ} \) has an exponent of 1.
- The concentration of \( I^{\circ} \) also has an exponent of 1.
- The concentration of \( OH^{\circ} \) is in the denominator, which implies a negative exponent of -1.

Therefore, the total order of the reaction can be calculated as:
\[ \text{Total Order} = 1 + 1 - 1 = 1 \]

2. **Identify the Molecularity of the Reaction:**
- Molecularity refers to the number of reactant molecules involved in the rate-determining step of the reaction.
- In this case, there are two reactants (\( I^{\circ} \) and \( OCl^{\circ} \)) participating in the reaction.

Thus, the molecularity of the reaction is:
\[ \text{Molecularity} = 2 \]

### Summary for Part A:
- **Order of the reaction:** 1
- **Molecularity of the reaction:** 2

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### Part B: Order and Molecularity with respect to \([OH^{\circ}]\)

1. **Identify the Order with respect to \([OH^{\circ}]\):**
- The order with respect to a specific reactant is determined by the exponent of that reactant in the rate law.
- In this case, the exponent of \([OH^{\circ}]\) in the rate law is -1.

Therefore, the order with respect to \([OH^{\circ}]\) is:
\[ \text{Order with respect to } [OH^{\circ}] = -1 \]

2. **Identify the Molecularity with respect to \([OH^{\circ}]\):**
- Molecularity is defined based on the reactants that are involved in the elementary step of the reaction.
- Since \([OH^{\circ}]\) is not a reactant in the reaction, its molecularity is considered to be zero.

Therefore, the molecularity with respect to \([OH^{\circ}]\) is:
\[ \text{Molecularity with respect to } [OH^{\circ}] = 0 \]

### Summary for Part B:
- **Order with respect to \([OH^{\circ}]\):** -1
- **Molecularity with respect to \([OH^{\circ}]\):** 0

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### Final Answers:
- **Part A:**
- Order: 1
- Molecularity: 2
- **Part B:**
- Order with respect to \([OH^{\circ}]\): -1
- Molecularity with respect to \([OH^{\circ}]\): 0

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