In a reaction `H_(2)+I_(2) to 2HI` the rate of disappearace of `I_(2)` is found to be `10^(-6)` mole per litre per second . What would be the corresponding rate of appearance of HI?

Explanation

To solve the problem, we need to relate the rate of disappearance of \( I_2 \) to the rate of appearance of \( HI \) using stoichiometry from the balanced chemical equation. The reaction is:

\[ H_2 + I_2 \rightarrow 2HI \]

### Step-by-step Solution:

1. **Identify the Stoichiometry of the Reaction:**
From the balanced equation, we can see that 1 mole of \( I_2 \) produces 2 moles of \( HI \). This means that the rate of disappearance of \( I_2 \) is related to the rate of appearance of \( HI \) by a factor of 2.
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