1. Tardigrade
  2. Question
  3. Chemistry
  4. For a particular reaction, Δ H°=-40 kJ and Δ S° =-200 J K -1 mol -1. The reaction is

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Q. For a particular reaction, $\Delta H^{\circ}=-40\, kJ$ and $\Delta S^{\circ}$ $=-200 \, J K ^{-1}\, mol ^{-1}$. The reaction is

Thermodynamics

Solution:

$\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$
When $\Delta G^{\circ}=0, T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}$
$=\frac{-40 \times 10^{3}}{-200}=200\, K$