For a gas phase reaction, $PCl_{5}(g)\rightleftharpoons PCl_{3}(g)+Cl_{2}(g)$ which of the following conditions are correct?

Q: For a gas phase reaction, $PCl_{5}(g)\rightleftharpoons PCl_{3}(g)+Cl_{2}(g)$ which of the following conditions are correct ?

For the gas phase reaction $ PCl_{5}(g) \rightleftharpoons PCl_{3}(g) + Cl_{2}(g) $, we need to analyze the thermodynamic properties, specifically the changes in enthalpy ($\Delta H$) and entropy ($\Delta S$). ### Explanation for Option 4: $\Delta H > 0, \Delta S > 0$ 1. **$\Delta H > 0$:** - The reaction involves the breaking of a bond in $ PCl_{5} $ to form $ PCl_{3} $ and $ Cl_{2} $. Breaking chemical bonds requires energy, which means the reaction is endothermic. Therefore, $\Delta H$ is positive. 2. **$\Delta S > 0$:** - Entropy ($\Delta S$) is a measure of disorder or randomness. In this reaction, one mole of $ PCl_{5} $ gas is converted into two moles of gas ($ PCl_{3} $ and $ Cl_{2} $). The increase in the number of gas molecules results in an increase in disorder, thus $\Delta S$ is positive. ### Explanation for why other options are incorrect: 1. **Option 1: $\Delta H 0, \Delta S 0$ correctly indicates an endothermic reaction, $\Delta S < 0$ is incorrect because the reaction results in more gas molecules, increasing entropy. 3. **Option 3: $\Delta H = 0, \Delta S < 0$** - $\Delta H = 0$ suggests no enthalpy change, which is incorrect as the reaction is endothermic. - $\Delta S 0$ and $\Delta S > 0$.

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For the gas phase reaction $ PCl_{5}(g) \rightleftharpoons PCl_{3}(g) + Cl_{2}(g) $, we need to analyze the thermodynamic properties, specifically the changes in enthalpy ($\Delta H$) and entropy ($\Delta S$). ### Explanation for Option 4: $\Delta H > 0, \Delta S > 0$ 1. **$\Delta H > 0$:** - The reaction involves the breaking of a bond in $ PCl_{5} $ to form $ PCl_{3} $ and $ Cl_{2} $. Breaking chemical bonds requires energy, which means the reaction is endothermic. Therefore, $\Delta H$ is positive. 2. **$\Delta S > 0$:** - Entropy ($\Delta S$) is a measure of disorder or randomness. In this reaction, one mole of $ PCl_{5} $ gas is converted into two moles of gas ($ PCl_{3} $ and $ Cl_{2} $). The increase in the number of gas molecules results in an increase in disorder, thus $\Delta S$ is positive. ### Explanation for why other options are incorrect: 1. **Option 1: $\Delta H < 0, \Delta S < 0$** - $\Delta H < 0$ suggests an exothermic reaction, which is incorrect because bond breaking in $ PCl_{5} $ is endothermic. - $\Delta S < 0$ implies a decrease in disorder, which is incorrect as the reaction increases the number of gas molecules, leading to increased disorder. 2. **Option 2: $\Delta H > 0, \Delta S < 0$** - While $\Delta H > 0$ correctly indicates an endothermic reaction, $\Delta S < 0$ is incorrect because the reaction results in more gas molecules, increasing entropy. 3. **Option 3: $\Delta H = 0, \Delta S < 0$** - $\Delta H = 0$ suggests no enthalpy change, which is incorrect as the reaction is endothermic. - $\Delta S < 0$ is incorrect for the same reason as above; the reaction increases the number of gas molecules, thus increasing entropy. In conclusion, Option 4 correctly describes the thermodynamic changes for the reaction $ PCl_{5}(g) \rightleftharpoons PCl_{3}(g) + Cl_{2}(g) $ with $\Delta H > 0$ and $\Delta S > 0$.