A first order reaction is carried out with an initial concentration of `10` mol litre and `80%` of the reactant changes into product in `10 sec`. Now if the same reaction is carried out with an intial concentration of `5` mol per litre the percentage of the reactant changing to the product in `10` sec is:

Options

A. `40`
B. `80`
C. `60`
D. `50`

Explanation

To solve the problem, we need to understand the characteristics of first-order reactions and how the change in concentration affects the percentage of reactant converted to product over a given time.

### Step-by-Step Solution:

1. **Understanding the Reaction Type**: We are dealing with a first-order reaction. For first-order reactions, the time required for a certain fraction of the reactant to convert to product is independent of the initial concentration.

2. **Initial Conditions**: The initial concentration of the reactant in the first scenario is given as `10 mol/L`, and it is stated that `80%` of this reactant converts to product in `10 seconds`.

3. **Calculating the Amount Converted**: If `80%` of `10 mol/L` reacts in `10 seconds`, then the amount that reacts can be calculated as:
\[ \text{Amount reacted} = 0.80 \times 10 \, \text{mol/L} = 8 \, \text{mol/L} \]
Therefore, the concentration of the reactant after `10 seconds` is:
\[ \text{Remaining concentration} = 10 \, \text{mol/L} - 8 \, \text{mol/L} = 2 \, \text{mol/L} \]

4. **New Initial Concentration**: Now, we are conducting the same reaction with an initial concentration of `5 mol/L`.

5. **Applying First-Order Reaction Characteristics**: Since the reaction is first-order, the percentage of reactant that reacts in a given time (in this case, `10 seconds`) remains constant regardless of the initial concentration.

6. **Calculating the Percentage for New Concentration**: Since the time is the same (`10 seconds`), and the reaction characteristics do not change, the percentage of reactant that converts to product will also be `80%` for the new initial concentration of `5 mol/L`.

7. **Conclusion**: Therefore, the percentage of the reactant that changes to product in `10 seconds` with an initial concentration of `5 mol/L` is:
\[ \text{Percentage converted} = 80\% \]

### Final Answer:
The percentage of the reactant changing to product in `10 seconds` with an initial concentration of `5 mol/L` is **80%**.

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