Calculate the Gibbs energy change for a reaction having $\Delta H = 31400 \ J$ and $\Delta S = 32 \ J \ K^{-1}$ at $1000^{\circ} C$. (in $J$)

  • A
    $-4668$
  • B
    $-9336$
  • C
    $-4073$
  • D
    $-2334$

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Similar Questions

For independent processes at $300 \ K$,determine the number of non-spontaneous processes from the following table:
Process $\Delta H \ (kJ \ mol^{-1})$ $\Delta S \ (J \ K^{-1} \ mol^{-1})$
$A$ $-25$ $-80$
$B$ $-22$ $40$
$C$ $25$ $-50$
$D$ $22$ $20$

For the reaction $2 \ NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$,when $\Delta S = -176.0 \ J \ K^{-1} \ mol^{-1}$ and $\Delta H = -57.8 \ kJ \ mol^{-1}$,the magnitude of $\Delta G$ at $298 \ K$ for the reaction is $...... \ kJ \ mol^{-1}$. (Nearest integer)

For a chemical reaction $A^{+} + B \rightleftharpoons C^{+} + D$ $(\Delta_{r} H^{0} = 80 \, kJ \, mol^{-1})$,the entropy change $\Delta_{r} S^{0}$ depends on the temperature $T$ (in $K$) as $\Delta_{r} S^{0} = 2T \, J \, K^{-1} \, mol^{-1}$.
Minimum temperature at which it will become spontaneous is ..... $K$. (Integer)

For vaporization of water at $1$ atmospheric pressure,the values of $\Delta H$ and $\Delta S$ are $40.63 \ kJ \ mol^{-1}$ and $108.8 \ J \ K^{-1} \ mol^{-1},$ respectively. The temperature when Gibb's energy change $(\Delta G)$ for this transformation will be zero,is ............ $K$. (in $.4$)

The temperature in $K$ at which $\Delta G = 0$, for a given reaction with $\Delta H = -20.5 \ kJ \ mol^{-1}$ and $\Delta S = -50.0 \ J \ K^{-1} \ mol^{-1}$ is:

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