The correct options are
(B) The cell reaction, $Pt (s)\left| H _2(g, 1 bar )\right| H ^{+}(a q, 0.01 M ) \| H ^{+}(a q, 0.1 M )\left| H _2(g, 1 bar )\right| Pt (s)$, is an entropy driven process.
(C) For racemization of an optically active compound, $\Delta S >0$.
(D) $\Delta S >0$, for $\left[ Ni \left( H _2 O \right)_6\right]^{2+}+3 en \rightarrow\left[ Ni ( en )_3\right]^{2+}+6 H _2 O$ (where en $=$ ethylenediamine).
To solve this problem, we need to evaluate each statement related to entropy (S) and determine the correct ones.
1. Analyzing Option A: For the reaction, \( \text{M(s)} + 2\text{H}^+(aq) \rightarrow \text{H}_2(g) + \text{M}^{2+}(aq) \), if \( \frac{dE_{\text{cell}}}{dT} = \frac{R}{F} \), then the entropy change of the reaction is \( R \) (assuming entropy and internal energy changes are temperature independent).
This statement is based on the assumption that entropy change is temperature independent, and the relationship between \( \frac{dE_{\text{cell}}}{dT} \) and \( \frac{R}{F} \) is correct. This statement is true.
2. Analyzing Option B: The cell reaction \( \text{Pt(s)} | \text{H}_2(g, 1 \, \text{bar}) | \text{H}^+(aq, 0.01 \, M) \| \text{H}^+(aq, 0.1 \, M) | \text{H}_2(g, 1 \, \text{bar}) | \text{Pt(s)} \), is an entropy-driven process.
This reaction involves a concentration difference in hydrogen ions and a change in the entropy of the system. This reaction is indeed entropy-driven due to the change in concentration and the corresponding entropy change. Therefore, this statement is true.
3. Analyzing Option C: For racemization of an optically active compound, \( \Delta S > 0 \).
Racemization of optically active compounds involves the conversion of one enantiomer into its mirror image. This process leads to an increase in disorder, which corresponds to a positive entropy change. Therefore, this statement is true.
4. Analyzing Option D: \( \Delta S > 0 \) for \( [\text{Ni(H}_2\text{O)}_6]^{2+} + 3 \, \text{en} \rightarrow [\text{Ni(en)}_3]^{2+} + 6 \, \text{H}_2\text{O} \) (where en = ethylenediamine).
In this reaction, the complex ion is formed by substituting water molecules with ethylenediamine molecules, which leads to an increase in the randomness of the system. This corresponds to a positive entropy change. Therefore, this statement is true.
Final Answer:
The correct options are B, C, D.
List-I (Symbol of electrical property) | List-II (Units) |
---|---|
A) \( \Omega \) | I) S cm\(^{-1}\) |
B) G | II) m\(^{-1}\) |
C) \( \kappa \) | III) S cm\(^2\) mol\(^{-1}\) |
D) G\(^*\) | IV) S |
Two identical concave mirrors each of focal length $ f $ are facing each other as shown. A glass slab of thickness $ t $ and refractive index $ n_0 $ is placed equidistant from both mirrors on the principal axis. A monochromatic point source $ S $ is placed at the center of the slab. For the image to be formed on $ S $ itself, which of the following distances between the two mirrors is/are correct: