For a certain thermochemical reaction $$M \rightarrow N$$ at $$T = 400$$ K, $$\Delta H^o = 77.2$$ kJ mol$$^{-1}$$, $$\Delta S^o = 122$$ J K$$^{-1}$$, log equilibrium constant ($$\log K$$) is $$-$$ _____ $$\times 10^{-1}$$.

We need to determine the value of the log equilibrium constant ($$\log K$$) for the given thermochemical reaction $$M \rightarrow N$$ at $$T = 400 \text{ K}$$.

The relationship between standard Gibbs free energy change ($$\Delta G^o$$), enthalpy change ($$\Delta H^o$$), and entropy change ($$\Delta S^o$$) is:

$$\Delta G^o = \Delta H^o - T\Delta S^o$$

Substituting the given values ($$\Delta H^o = 77.2 \text{ kJ mol}^{-1} = 77200 \text{ J mol}^{-1}$$, $$T = 400 \text{ K}$$, and $$\Delta S^o = 122 \text{ J K}^{-1} \text{ mol}^{-1}$$):

$$\Delta G^o = 77200 - (400 \times 122)$$

$$\Delta G^o = 77200 - 48800 = 28400 \text{ J mol}^{-1}$$

The relationship between standard Gibbs free energy change ($$\Delta G^o$$) and the equilibrium constant ($$K$$) is:

$$\Delta G^o = -2.303 R T \log K$$

Rearranging the formula to solve for $$\log K$$ (using $$R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$$):

$$\log K = -\frac{\Delta G^o}{2.303 R T}$$

$$\log K = -\frac{28400}{2.303 \times 8.314 \times 400}$$

$$\log K = -\frac{28400}{7658.84}$$

$$\log K \approx -3.708$$

$$-3.708 = -37.08 \times 10^{-1}$$

Rounding to the nearest integer, the value inside the blank is $$37$$.

The answer is 37.