For the reaction
`H_(2) (g)+I_(2)(g) rarr 2HI(g)`, the rate of disappearance of `H_(2)` is `1.0 xx 10^(-4) mol L^(-1) s^(-1)`. The rate of appearance of `HI` will be

Options

A. `1.0 xx 10^(-4) mol L^(-1) s^(-1)`
B. `2.0 xx 10^(-4) mol L^(-1) s^(-1)`
C. `0.50 xx 10^(-4) mol L^(-1) s^(-1)`
D. `4.0 xx 10^(-4) mol L^(-1) s^(-1)`

Explanation

To solve the problem, we need to determine the rate of appearance of HI for the reaction:

\[ H_2(g) + I_2(g) \rightarrow 2HI(g) \]

Given that the rate of disappearance of \( H_2 \) is \( 1.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \), we can use the stoichiometry of the reaction to find the rate of appearance of \( HI \).

### Step-by-Step Solution:

1. **Write the Rate of Reaction**:
The rate of reaction can be expressed in terms of the change in concentration of the reactants and products. For the given reaction, we can write:
\[ \text{Rate of reaction} = -\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt} \]

2. **Relate the Rates**:
From the stoichiometry of the reaction, we can relate the rate of disappearance of \( H_2 \) to the rate of appearance of \( HI \):
\[ -\frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt} \]

3. **Substitute the Given Rate**:
We know that the rate of disappearance of \( H_2 \) is given as:
\[ -\frac{d[H_2]}{dt} = 1.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]
Therefore, we can substitute this value into the equation:
\[ 1.0 \times 10^{-4} = \frac{1}{2} \frac{d[HI]}{dt} \]

4. **Solve for the Rate of Appearance of HI**:
To find \( \frac{d[HI]}{dt} \), we can rearrange the equation:
\[ \frac{d[HI]}{dt} = 2 \times 1.0 \times 10^{-4} \]
\[ \frac{d[HI]}{dt} = 2.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Final Answer:
The rate of appearance of \( HI \) is:
\[ \frac{d[HI]}{dt} = 2.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

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