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In which case change in entropy is negat...

In which case change in entropy is negative?

A

Sublimation of solid to gas

B

`2H(g) to H_2(g)`

C

Evaporation of water

D

Expansion of a gas at constant temperature.

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The correct Answer is:
To determine in which case the change in entropy is negative, we need to analyze each option provided in the question. ### Step-by-Step Solution: 1. **Understanding Entropy**: - Entropy (S) is a measure of the randomness or disorder of a system. A negative change in entropy (ΔS < 0) indicates a decrease in disorder. 2. **Analyzing Each Option**: - **Option A: Sublimation of solid to gas**: - In sublimation, a solid transforms directly into a gas. This process increases the randomness as gas molecules have higher kinetic energy and more freedom of movement compared to solids. Therefore, ΔS > 0. This option is not correct. - **Option B: 2 H atoms in gaseous state to form H2 gaseous**: - Here, we have two gaseous hydrogen atoms combining to form a single molecule of hydrogen gas (H2). The number of gaseous moles decreases from 2 (reactants) to 1 (product). - The change in the number of gaseous moles (ΔnG) = 1 (product) - 2 (reactants) = -1. - A decrease in the number of gaseous moles indicates a decrease in randomness, thus ΔS < 0. This option is correct. - **Option C: Evaporation of water**: - During evaporation, liquid water molecules transition to the gas phase. This process increases disorder because gas molecules have more freedom and energy than liquid molecules. Therefore, ΔS > 0. This option is not correct. - **Option D: Expansion of gas at constant temperature**: - When a gas expands at constant temperature, its volume increases, leading to greater randomness and disorder among the gas molecules. Thus, ΔS > 0. This option is not correct. 3. **Conclusion**: - After analyzing all options, we find that the only case where the change in entropy is negative is **Option B**, where 2 H atoms in gaseous state combine to form H2 gaseous. ### Final Answer: The change in entropy is negative in **Option B: 2 H atoms in gaseous state to form H2 gaseous**. ---

To determine in which case the change in entropy is negative, we need to analyze each option provided in the question. ### Step-by-Step Solution: 1. **Understanding Entropy**: - Entropy (S) is a measure of the randomness or disorder of a system. A negative change in entropy (ΔS < 0) indicates a decrease in disorder. 2. **Analyzing Each Option**: ...
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For a sponaneous reaction, the free energy change must be negative, Delta G=Delta H-T Delta S, Delta H is the enthalpy change during the reaction. T is the absolute temperature, and Delta S is the change in entropy during the reaction. Consider a reaction such as the formation of an oxide M+O_(2) to MO Dioxygen is used up in the course of this reaction. Gases have a more random structure (less ordered) than liquid or solids. Consequently gases have a higher entropy than liquids and solids. In this reaction S (entropy or randomness) decreases, hence Delta S is negative. Thus, if the temperature is raised then T Delta S becomes more negative,Since, TDelta S is substracted in the equation, then Delta G becomes less negative. Thus, the free energy change increases with the increase in temperature. The free energy changes that occur when one mole of common reactant (in this case dioxygen) is used may be plotted graphically aginst temperature for a number of reactions of metals to their oxides. The following plot is called an Ellingham diagram for metal oxide. Understanding of Ellingham diagram is extremely important for the efficient extraction of metals. As per the Ellingham diagram of oxides which of the following conclusion is true ?

The change in entropy, Delta S is positive for an endothermic reaction, if enthalpy change Delta H occurs at the same temperature T, then the reaction is feasible

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