The correct option(s) about entropy $(S)$ is(are)
$[R =$ gas constant, $F =$ Faraday constant, $T =$ Temperature $]$
$(A)$ For the reaction, $M_{(s)} + 2H^{+}_{(aq)} \rightarrow H_{2(g)} + M^{2+}_{(aq)}$, if $\frac{dE_{cell}}{dT} = \frac{R}{F}$, then the entropy change of the reaction is $R$ (assume that entropy and internal energy changes in entropy and internal energy are temperature independent).
$(B)$ The cell reaction, $Pt_{(s)} \mid H_2(g, 1 \ bar) \mid H^{+}(aq, 0.01 \ M) \parallel H^{+}(aq, 0.1 \ M) \mid H_2(g, 1 \ bar) \mid Pt_{(s)}$, is an entropy driven process.
$(C)$ For racemization of an optically active compound, $\Delta S > 0$.
$(D)$ $\Delta S > 0$, for $[Ni(H_2O)_6]^{2+} + 3en \rightarrow [Ni(en)_3]^{2+} + 6H_2O$ (where $en =$ ethylenediamine).

  • A
    $B, C, D$
  • B
    $B, C$
  • C
    $B, D$
  • D
    $A, B$

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Similar Questions

Select the incorrect statement.

The given complex $M(ABCDEF)$ exhibits which of the following isomerism?

$0.02 \, \text{mole}$ of $[Co(NH_3)_5Br]Cl_2$ and $0.02 \, \text{mole}$ of $[Co(NH_3)_5Cl]SO_4$ are present in $200 \, \text{cc}$ of a solution $X$. The number of moles of the precipitates $Y$ and $Z$ that are formed when the solution $X$ is treated with excess silver nitrate and excess barium chloride are respectively:

Match each coordination compound in List-$I$ with an appropriate pair of characteristics from List-$II$ and select the correct answer using the code given below the lists.
$\{ en = H_2NCH_2CH_2NH_2 \}$; atomic numbers: $\{Ti = 22; Cr = 24; Co = 27; Pt = 78\}$
List-$I$ List-$II$
$P.$ $[Cr(NH_3)_4Cl_2]Cl$ $1.$ Paramagnetic and exhibits ionisation isomerism
$Q.$ $[Ti(H_2O)_5Cl](NO_3)_2$ $2.$ Diamagnetic and exhibits cis-trans isomerism
$R.$ $[Pt(en)(NH_3)_2Cl_2]NO_3$ $3.$ Paramagnetic and exhibits cis-trans isomerism
$S.$ $[Co(NH_3)_4(NO_3)_2]NO_3$ $4.$ Diamagnetic and exhibits ionisation isomerism
Codes: $P, Q, R, S$

$1\,L, 0.02\,M$ solution of $[Co(NH_3)_5SO_4]Br$ is mixed with $1\,L, 0.02\,M$ solution of $[Co(NH_3)_5Br]SO_4$. The resulting solution is divided into two equal parts $(X)$ and treated with excess $AgNO_3$ solution and $BaCl_2$ solution respectively as shown below:
$1\,L$ Solution $(X) + AgNO_3$ solution (excess) $\rightarrow Y$
$1\,L$ Solution $(X) + BaCl_2$ solution (excess) $\rightarrow Z$
The number of moles of $Y$ and $Z$ respectively are:

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