For a reaction to be spontaneous at all temperatures, the change in Gibbs free energy (\( \Delta G \)) must be negative for all temperatures.
The expression for \( \Delta G \) is: \[ \Delta G = \Delta H - T\Delta S \] For the reaction to be spontaneous at all temperatures, \( \Delta G \) should be negative. This will happen if: \[ \Delta_r H<0 \quad {and} \quad \Delta_r S>0 \] Thus, option (B) is the correct answer.
Which of the following changes alone would cause increase in the value of the equilibrium constant of the reaction?
\(PCl_5(g) \rightarrow PCl_3(g) + Cl_2(g); \Delta H>0\)
For the reaction:
$3Fe_{(s)} + 2O_2{(g)} \rightarrow Fe_3O_4{(s)}$
$\Delta H = -1650\,\text{kJ mol}^{-1}$, $\Delta S = -600\,\text{J K}^{-1} \text{mol}^{-1}$ at $300\,\text{K}$. What is the value of free energy change for the reaction at $300\,\text{K}$?
Kepler's second law (law of areas) of planetary motion leads to law of conservation of