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(A) $(I)$ The total entropy change $(\Delta S_{	ext{total}})$ is given by $\Delta S_{	ext{total}} = \Delta S_{	ext{sys}} + \Delta S_{	ext{surr}}$.

Vedclass · Accepted Answer

$I, III$ only

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(A) $(I)$ The total entropy change $(\Delta S_{	ext{total}})$ is given by $\Delta S_{	ext{total}} = \Delta S_{	ext{sys}} + \Delta S_{	ext{surr}}$. Therefore,$\Delta S_{	ext{sys}} = \Delta S_{	ext{total}} - \Delta S_{	ext{surr}}$. The statement $I$ is incorrect.$(II)$ For the process $A_{(l)} ightarrow A_{(s)}$,the system transitions from a liquid to a solid state. Since the disorder decreases,the entropy change is negative (entropy decreases). The statement $II$ is correct.$(III)$ The $SI$ unit of entropy $(S)$ is $J K^{-1} mol^{-1}$. The statement $III$ is correct.

Identify the incorrect statements from the following:
$I. \quad \Delta S_{\text{system}} = (\Delta S_{\text{total}} + \Delta S_{\text{surr.}})$
$II. A_{(l)} \rightarrow A_{(s)}$: for this process entropy change decreases
$III.$ Entropy units are $J K^{-1} mol^{-1}$

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Identify the incorrect statements from the following:$I. \quad \Delta S_{\text{system}} = (\Delta S_{\text{total}} + \Delta S_{\text{surr.}})$$II. A_{(l)} \rightarrow A_{(s)}$: for this process entropy change decreases$III.$ Entropy units are $J K^{-1} mol^{-1}$