The change in internal energy for the reaction $$\text{H}_2(g) + \text{Br}_2(g) \rightarrow 2\text{HBr}(l)$$ when 2.0 moles each of $$\text{Br}_2(g)$$ and $$\text{H}_2(g)$$ react is, given that for $$\text{H}_2(g) + \text{Br}_2(g) \rightarrow 2\text{HBr}(g)$$ the enthalpy of reaction is $$-109$$ kJ and the enthalpy of vaporisation of HBr is 213 kJ $$\text{mol}^{-1}$$

A

$$-644$$ kJ

B

644 kJ

C

$$-322$$ kJ

D

$$-1070$$ kJ