Question 6.18 For the reaction,
2Cl(g)→Cl2(g), what are the signs of ΔH and ΔS ?
Class XI Thermodynamics Page 183
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Answered by
56
2Cl (g) ------->Cl₂(g)
in the above reaction, a molecule Cl₂ forms from its two Cl atoms and energy is the released for the formation of Cl₂ bond. hence, ΔH is negative, in this reaction randomness ( entropy) also decreases because 2 mol of Cl atoms have more randomness than 1 mol of Cl₂ molecules. hence, ΔS = -ve
in the above reaction, a molecule Cl₂ forms from its two Cl atoms and energy is the released for the formation of Cl₂ bond. hence, ΔH is negative, in this reaction randomness ( entropy) also decreases because 2 mol of Cl atoms have more randomness than 1 mol of Cl₂ molecules. hence, ΔS = -ve
kvnmurty:
excellent explanation.
Answered by
11
Hi
Here is your answer,
For an isolated system, ΔU = 0 and for a spontaneous process, total entropy change must be positive. Like for example :- Consider the diffusion of two gases A and B into each in a closed container which is isolated from the surrounding. The two gases A and B are separated by a movable partition. When partition is removed , the gases begin to diffuse into each other and the system becomes more disordered. It shows that ΔS > 0 and ΔU = 0 for this process.
Moreover, ΔS = qrev/T = ΔU +pΔV/T = pΔV/T
TΔS or ΔS > 0
Hope it helps you !
Here is your answer,
For an isolated system, ΔU = 0 and for a spontaneous process, total entropy change must be positive. Like for example :- Consider the diffusion of two gases A and B into each in a closed container which is isolated from the surrounding. The two gases A and B are separated by a movable partition. When partition is removed , the gases begin to diffuse into each other and the system becomes more disordered. It shows that ΔS > 0 and ΔU = 0 for this process.
Moreover, ΔS = qrev/T = ΔU +pΔV/T = pΔV/T
TΔS or ΔS > 0
Hope it helps you !
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