Decomposition of `Hl (g)` on Gold surface is zero order reaction. Initially, few moles of `H_(2)` are present in container then which of the following graph is correct ?

Options

A.
B.
C.
D.

Explanation

To solve the question regarding the decomposition of HI (g) on a gold surface, which is a zero-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Zero-Order Reactions**:
- In a zero-order reaction, the rate of reaction is constant and does not depend on the concentration of the reactants. The rate law can be expressed as:
\[ \text{Rate} = k \]
- The integrated rate law for a zero-order reaction is given by:
\[ [A] = [A]_0 - kt \]
where \([A]\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.

2. **Graphical Representation**:
- For a zero-order reaction, if we plot the concentration of the reactant (in this case, HI) against time, the graph will be a straight line with a negative slope. This indicates that the concentration decreases linearly over time.

3. **Initial Concentration of H2**:
- The problem states that initially, there are a few moles of \(H_2\) present. This means that at \(t = 0\), the concentration of \(H_2\) is not zero.

4. **Analyzing the Options**:
- Since the graph of a zero-order reaction will show a linear decrease in concentration, we need to identify which graph correctly represents this behavior while also accounting for the initial concentration of \(H_2\).
- The options presented likely include graphs that either start from zero or have a negative slope. Since we have an initial concentration of \(H_2\), the graph must start from a positive value and decrease over time.

5. **Conclusion**:
- The correct graph will be the one that starts from a positive concentration of \(H_2\) and shows a linear decrease over time. Therefore, the correct option is **B**, which represents the behavior of a zero-order reaction with an initial concentration of \(H_2\).

### Final Answer:
The correct graph is option **B**.