For the reaction $$4\text{NO}_2(g) + \text{O}_2(g) \to 2\text{N}_2\text{O}_5(g)$$, $$\Delta H_{reaction} = -112\ \text{kJ}$$. If the $$\text{N}_2\text{O}_5$$ is assumed to be formed in the reaction as a solid, $$\Delta H_{reaction}$$ will be (the enthalpy of sublimation of $$\text{N}_2\text{O}_5$$ is $$54\ \text{kJ mol}^{-1}$$)

A

$$-220\ \text{kJ}$$

B

$$-4\ \text{kJ}$$

C

$$-166\ \text{kJ}$$

D

$$-332\ \text{kJ}$$