For which of the following reactions, $$\Delta H$$ is equal to $$\Delta U$$?

We begin by recalling the thermodynamic relation that connects the enthalpy change $$\Delta H$$ with the internal energy change $$\Delta U$$ for any chemical reaction carried out at constant temperature:

$$\Delta H \;=\; \Delta U \;+\; \Delta n_g \, R T$$

Here, $$\Delta n_g$$ represents the difference between the total number of moles of gaseous products and the total number of moles of gaseous reactants, $$R$$ is the universal gas constant, and $$T$$ is the absolute temperature. It is evident from this formula that $$\Delta H$$ will be numerically equal to $$\Delta U$$ if and only if $$\Delta n_g = 0$$, because then the extra term $$\Delta n_g R T$$ vanishes.

We shall therefore calculate $$\Delta n_g$$ for each reaction listed.

Option A: $$\mathrm{N_2(g) + 3\,H_2(g) \;\rightarrow\; 2\,NH_3(g)}$$

Total moles of gaseous reactants $$= 1 + 3 = 4$$.
Total moles of gaseous products $$= 2$$.
So, $$\Delta n_g = 2 - 4 = -2 \neq 0$$. Hence $$\Delta H \neq \Delta U$$ for Option A.

Option B: $$\mathrm{2\,HI(g) \;\rightarrow\; H_2(g) + I_2(g)}$$

Total moles of gaseous reactants $$= 2$$.
Total moles of gaseous products $$= 1 + 1 = 2$$.
Therefore, $$\Delta n_g = 2 - 2 = 0$$. Because the difference in gaseous moles is zero, the term $$\Delta n_g R T$$ is zero, giving $$\Delta H = \Delta U$$ for Option B.

Option C: $$\mathrm{2\,SO_2(g) + O_2(g) \;\rightarrow\; 2\,SO_3(g)}$$

Total moles of gaseous reactants $$= 2 + 1 = 3$$.
Total moles of gaseous products $$= 2$$.
Thus, $$\Delta n_g = 2 - 3 = -1 \neq 0$$. Consequently, $$\Delta H \neq \Delta U$$ for Option C.

Option D: $$\mathrm{2\,NO_2(g) \;\rightarrow\; N_2O_4(g)}$$

Total moles of gaseous reactants $$= 2$$.
Total moles of gaseous products $$= 1$$.
Hence, $$\Delta n_g = 1 - 2 = -1 \neq 0$$, and $$\Delta H \neq \Delta U$$ for Option D.

Among all the alternatives, only Option B yields $$\Delta n_g = 0$$, fulfilling the condition for $$\Delta H$$ to equal $$\Delta U$$.

Hence, the correct answer is Option B.