If the equilibrium constant of a process is $3.8 \times 10^{-3}$ at $25^{\circ} C$, what is the standard free energy change of the process? $(R = 8.314 \ J \ mol^{-1} \ K^{-1}, \log 0.0038 = -2.42)$

For a hypothetical reversible reaction $\frac{1}{2} A_{2(g)} + \frac{3}{2} B_{2(g)} \rightarrow AB_{3(g)}$; $\Delta H = -20 \, kJ$. If the standard entropies of $A_2, B_2$ and $AB_3$ are $60, 40$ and $50 \, J K^{-1} mol^{-1}$ respectively,at what temperature $(K)$ will the reaction be in equilibrium?

The free energy change for a reversible reaction at equilibrium is

Find the equilibrium constant for the reaction below at $25$ $^{\circ}C$:
$H_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons H_{2}O_{(g)}$ [ $\Delta_{f}G^{o} = -54.64 \ kcal$ ].

Assertion: For every chemical reaction at equilibrium,the standard Gibbs energy change is zero.
Reason: At constant temperature and pressure,a chemical reaction is spontaneous in the direction of decreasing Gibbs energy.

The value of $\Delta G^{\ominus}$ for the phosphorylation of glucose in glycolysis is $13.8 \, kJ \, mol^{-1}$. Find the value of $K_{c}$ at $298 \, K$.