For the reaction $$H_2F_2(g) \rightarrow H_2(g) + F_2(g)$$
$$\Delta U = -59.6 \text{ kJ mol}^{-1}$$ at $$27°C$$
The enthalpy change for the above reaction is $$-$$ ______ $$\text{kJ mol}^{-1}$$ (nearest integer) (Given: $$R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$$)

We are given: $$H_2F_2(g) \rightarrow H_2(g) + F_2(g)$$ and $$\Delta U = -59.6 \text{ kJ mol}^{-1}$$ at 27°C, and we need to find the enthalpy change $$\Delta H$$.

Since the relation between $$\Delta H$$ and $$\Delta U$$ is given by $$\Delta H = \Delta U + \Delta n_g RT$$, where $$\Delta n_g$$ is the change in the number of moles of gaseous products minus gaseous reactants, we proceed to determine $$\Delta n_g$$.

Substituting the stoichiometric coefficients shows that the moles of gaseous products are 1 ($$H_2$$) + 1 ($$F_2$$) = 2 and the moles of gaseous reactants are 1 ($$H_2F_2$$), which gives $$\Delta n_g = 2 - 1 = 1$$.

At 27°C (T = 300 K) and with R = 8.314 J K^{-1} mol^{-1}, we calculate $$\Delta n_g RT = 1 \times 8.314 \times 300 = 2494.2 \text{ J mol}^{-1} = 2.4942 \text{ kJ mol}^{-1}$$.

Therefore, $$\Delta H = \Delta U + \Delta n_g RT = -59.6 + 2.4942 = -57.1 \text{ kJ mol}^{-1}$$.

The question asks for the magnitude (with the negative sign already indicated): - ____ kJ mol$$^{-1}$$. Rounding to the nearest integer gives $$|\Delta H| \approx 57 \text{ kJ mol}^{-1}$$.

Therefore, the answer is 57.