The heats of combustion of $C_{(s)}$,$H_{2(g)}$,and $C_{2}H_{6(g)}$ are $-x_{1}$,$-x_{2}$,and $-x_{3}$ respectively. The heat of formation of $C_{2}H_{6(g)}$ is:

  • A
    $-x_{1} - x_{2} + x_{3}$
  • B
    $-2x_{1} - 3x_{2} + x_{3}$
  • C
    $x_{1} + x_{2} - x_{3}$
  • D
    $-x_{3} + 2x_{1} + 3x_{2}$

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Match the transformations in Column-$I$ with the appropriate options in Column-$II$.
Column-$I$ Column-$II$
$(A) \; CO_{2(s)} \to CO_{2(g)}$ $(p) \; \text{Transition state}$
$(B) \; CaCO_{3(s)} \to CaO_{(s)} + CO_{2(g)}$ $(q) \; \text{Allotropic change}$
$(C) \; 2H^{\cdot} \to H_{2(g)}$ $(r) \; \Delta H > 0$
$(D) \; P_{\text{(white solid)}} \to P_{\text{(red solid)}}$ $(s) \; \Delta S > 0$
$(t) \; \Delta S < 0$

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For the reaction $2Cl_{(g)} \rightarrow Cl_{2(g)}$,the signs of $\Delta H$ and $\Delta S$ are respectively:

For a dimerization reaction,$2 A_{(g)} \rightarrow A_{2(g)}$ at $298 \ K$,$\Delta U^{\ominus} = -20 \ kJ \ mol^{-1}$,$\Delta S^{\ominus} = -30 \ J \ K^{-1} \ mol^{-1}$,then the $\Delta G^{\ominus}$ will be........ $J$

For the reaction
$A_{(\ell)} \rightarrow 2 B_{(g)}$
$\Delta U = 2.1 \; kcal, \Delta S = 20 \; cal \; K^{-1} \; mol^{-1}$ at $300 \; K$
Hence $\Delta G$ in $kcal \; mol^{-1}$ is

In an irreversible process taking place at constant $T$ and $P$ and in which only pressure-volume work is being done,the change in Gibbs free energy $(dG)$ and change in entropy $(dS)$ satisfy the criteria:

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