The entropy change associated with the conversion of $1 \ kg$ of ice at $273 \ K$ to water vapours at $383 \ K$ is: (Specific heat of water liquid and water vapours are $4.2 \ kJ \ K^{-1} \ kg^{-1}$ and $2.0 \ kJ \ K^{-1} \ kg^{-1},$ heat of fusion and vaporisation of water are $334 \ kJ \ kg^{-1}$ and $2491 \ kJ \ kg^{-1},$ respectively) $(\ln \ 273 = 5.61, \ln \ 373 = 5.92, \ln \ 383 = 5.95)$

  • A
    $7.90$
  • B
    $2.64$
  • C
    $8.49$
  • D
    $9.26$

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The free energy change for the following reactions are given below:
$C_2H_{2(g)} + \frac{5}{2}O_{2(g)} \to 2CO_{2(g)} + H_2O_{(l)}; \Delta G^o = -1234 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta G^o = -394 \ kJ$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}; \Delta G^o = -237 \ kJ$
What is the standard free energy change for the reaction $H_{2(g)} + 2C_{(s)} \to C_2H_{2(g)}$ in $kJ$?

Benzene burns according to the following equation at $300 \ K$ $(R = 8.314 \ J \ mol^{-1} K^{-1})$: $2 C_6H_{6(l)} + 15 O_{2(g)} \to 12 CO_{2(g)} + 6 H_2O_{(l)}$,$\Delta H^o = -6542 \ kJ/mol$. What is the $\Delta E^o$ for the combustion of $1.5 \ mol$ of benzene in $kJ$?

For the reaction $\frac{1}{2}X_2 + \frac{3}{2}Y_2 \to XY_3$,$\Delta H = -30 \ kJ/mol$. Given $\Delta S_{X_2} = 60 \ J/mol \cdot K$,$\Delta S_{Y_2} = 40 \ J/mol \cdot K$,and $\Delta S_{XY_3} = 50 \ J/mol \cdot K$,calculate the temperature at equilibrium in $K$.

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The enthalpy of neutralisation of $KOH$ by a weak acid $(HA)$ is $-12.2 \ kcal \ mol^{-1}$. If the weak acid is $10 \%$ ionised,then the enthalpy of dissociation of $1 \ mole$ of $HA$ is. Given that the enthalpy of neutralisation of $KOH$ by $HCl$ is $-13.7 \ kcal \ mol^{-1}$.

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If $\Delta G$ and $\Delta S$ for the reaction $A_{(g)} \rightarrow B_{(g)} + 2C_{(g)}$ at $2000 \ K$ are $-40 \ kJ \ mol^{-1}$ and $0.22 \ kJ \ K^{-1} \ mol^{-1}$ respectively,the change in internal energy for the same reaction approximately (in $kJ \ mol^{-1}$) is

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