For the reaction: $Cl_{2}(g)\rightarrow2Cl(g)$ the signs of $\Delta H$ and $\Delta S$ are:

Q: For the reaction: $Cl_{2}(g)\rightarrow2Cl(g)$ the signs of $\Delta H$ and $\Delta S$ are:

For the reaction $ Cl_{2}(g) \rightarrow 2Cl(g) $, we need to determine the signs of $\Delta H$ (change in enthalpy) and $\Delta S$ (change in entropy). 1. **Explanation for Option 2 (Both positive):** - **$\Delta H$ (Enthalpy Change):** The reaction involves breaking a chlorine molecule ($Cl_{2}$) into two chlorine atoms ($2Cl$). Breaking chemical bonds requires energy input, which means the process is endothermic. Therefore, $\Delta H$ is positive because energy is absorbed to break the $Cl-Cl$ bond. - **$\Delta S$ (Entropy Change):** Entropy is a measure of disorder or randomness. In this reaction, one molecule of $Cl_{2}$ gas is dissociated into two separate $Cl$ atoms. The number of gas particles increases, leading to greater disorder. Thus, $\Delta S$ is positive because the system becomes more disordered. Therefore, both $\Delta H$ and $\Delta S$ are positive, supporting Option 2 as the correct answer. 2. **Explanation for why other options are incorrect:** - **Option 1 (Both negative):** This option suggests that both $\Delta H$ and $\Delta S$ are negative. However, as explained, breaking bonds requires energy (positive $\Delta H$), and the increase in the number of particles increases disorder (positive $\Delta S$). Therefore, both cannot be negative. - **Option 3 ($\Delta H$ negative and $\Delta S$ positive):** This option suggests that the reaction is exothermic ($\Delta H$ negative), which is incorrect because breaking bonds requires energy input, making $\Delta H$ positive. - **Option 4 ($\Delta H$ positive and $\Delta S$ negative):** While $\Delta H$ is correctly identified as positive, $\Delta S$ cannot be negative because the reaction results in an increase in the number of gas particles, leading to increased disorder and a positive $\Delta S$. In conclusion, Option 2 is the correct answer because both $\Delta H$ and $\Delta S$ are positive for the given reaction.

Question

Accepted Answer

For the reaction $ Cl_{2}(g) \rightarrow 2Cl(g) $, we need to determine the signs of $\Delta H$ (change in enthalpy) and $\Delta S$ (change in entropy).

1. **Explanation for Option 2 (Both positive):**

- **$\Delta H$ (Enthalpy Change):** The reaction involves breaking a chlorine molecule ($Cl_{2}$) into two chlorine atoms ($2Cl$). Breaking chemical bonds requires energy input, which means the process is endothermic. Therefore, $\Delta H$ is positive because energy is absorbed to break the $Cl-Cl$ bond.

- **$\Delta S$ (Entropy Change):** Entropy is a measure of disorder or randomness. In this reaction, one molecule of $Cl_{2}$ gas is dissociated into two separate $Cl$ atoms. The number of gas particles increases, leading to greater disorder. Thus, $\Delta S$ is positive because the system becomes more disordered.

Therefore, both $\Delta H$ and $\Delta S$ are positive, supporting Option 2 as the correct answer.

2. **Explanation for why other options are incorrect:**

- **Option 1 (Both negative):** This option suggests that both $\Delta H$ and $\Delta S$ are negative. However, as explained, breaking bonds requires energy (positive $\Delta H$), and the increase in the number of particles increases disorder (positive $\Delta S$). Therefore, both cannot be negative.

- **Option 3 ($\Delta H$ negative and $\Delta S$ positive):** This option suggests that the reaction is exothermic ($\Delta H$ negative), which is incorrect because breaking bonds requires energy input, making $\Delta H$ positive.

- **Option 4 ($\Delta H$ positive and $\Delta S$ negative):** While $\Delta H$ is correctly identified as positive, $\Delta S$ cannot be negative because the reaction results in an increase in the number of gas particles, leading to increased disorder and a positive $\Delta S$.

In conclusion, Option 2 is the correct answer because both $\Delta H$ and $\Delta S$ are positive for the given reaction.