Deduce the rate law for the converison of `H_(2)` and `I_(2)` to `HI` at `440^(@)C` corresponding to the following proposed mechanism:
step I: `I(2) hArr 2I^-1 ("fast") (k_(1))`
Step II: `I^-1+H_(2) hArr IH_(2) ("fast") (k_(2))`
Step III: `IH_(2) + I^-1 rarr 2HI` (slow) `(k)`

Explanation

To deduce the rate law for the conversion of \( H_2 \) and \( I_2 \) to \( HI \) at \( 440^\circ C \) based on the proposed mechanism, we will analyze each step of the mechanism and derive the overall rate law.

### Step-by-Step Solution:

1. **Identify the Steps of the Mechanism:**
- **Step I:** \( I_2 \rightleftharpoons 2I^- \) (fast, rate constant \( k_1 \))
- **Step II:** \( I^- + H_2 \rightleftharpoons IH_2 \) (fast, rate constant \( k_2 \))
- **Step III:** \( IH_2 + I^- \rightarrow 2HI \) (slow, rate constant \( k \))
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