Solid $KClO_4$ is taken in a container maintained at a constant pressure of $1 \, atm$. Upon heating,the following equilibrium is obtained:
$2KClO_{4(s)} \rightleftharpoons 2KCl_{(s)} + 3O_{2(g)}$
If $\Delta H^o = 25 \, kcal/mol$ and $\Delta S^o = 50 \, cal/K \cdot mol$,at what temperature will equilibrium be established in the container (in $, K$)? (Ignore variation of $\Delta H^o$ and $\Delta S^o$ with temperature.)

  • A
    $298$
  • B
    $500$
  • C
    $5000$
  • D
    $300$

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Which of the following equations relates the temperature of a reaction with $\Delta H^{\circ}$ and $\Delta S^{\circ}$ at equilibrium?

Consider the reaction $X \rightleftharpoons Y$ at $300 \text{ K}$. If $\Delta H^\circ$ and $K$ are $28.40 \text{ kJ mol}^{-1}$ and $1.8 \times 10^{-7}$ at the same temperature,then the magnitude of $\Delta S^\circ$ for the reaction in $\text{J K}^{-1} \text{ mol}^{-1}$ is . . . . . . . (Nearest integer) (Given: $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$,$\ln 10 = 2.3$,$\log 3 = 0.48$,$\log 2 = 0.30$)

The reaction,$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$ is in equilibrium in a closed vessel at $298 \, K$. The partial pressure (in $atm$) of $CO_{2(g)}$ in the reaction vessel is closest to $....$ [Given : The change in Gibbs energies of formation at $298 \, K$ and $1 \, bar$ for $CaO_{(s)} = -603.501 \, kJ \, mol^{-1}$,$CO_{2(g)} = -394.389 \, kJ \, mol^{-1}$,$CaCO_{3(s)} = -1128.79 \, kJ \, mol^{-1}$,Gas constant,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$]

At $300 \ K$,for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,the equilibrium constant $K_p = 1.8 \times 10^{-7}$. Calculate its standard Gibbs free energy change $\Delta G^0$.

The correct relation is:

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