The standard free energy change $(\Delta G^{\circ})$ for $50\%$ dissociation of $N_2O_4$ into $NO_2$ at $27^{\circ}C$ and $1\,atm$ pressure is $-x\,J\,mol^{-1}$. The value of $x$ is $......$ (Nearest Integer)
[Given: $R = 8.31\,J\,K^{-1}\,mol^{-1}$,$\log 1.33 = 0.1239$,$\ln 10 = 2.3$]

  • A
    $520$
  • B
    $430$
  • C
    $931$
  • D
    $710$

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Similar Questions

For the reversible reaction,$A_{(s)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$,$\Delta G^{\circ} = -350 \ kJ$,which one of the following statements is true?

For the reaction $CH_{4(g)} + H_{2(g)} \longrightarrow C_2H_{6(g)}$,$K_p = 3.356 \times 10^{17}$,calculate $\Delta G^{\circ}$ for the reaction at $298 \ K$.

The equilibrium constant $(K)$ of a reaction may be written as :

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,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$]

The value of $\log _{10} K$ for a reaction $A \rightleftharpoons B$ is
(Given : $\Delta _{r} H_{298 K}^{\circ} = -54.07 \ kJ \ mol^{-1}$,$\Delta _{r} S_{298 K}^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$; $2.303 \times 8.314 \times 298 = 5705$)

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