We know that

$$\Delta $$H = $$\Delta $$E + $$\Delta $$n_gRT

In the reaction, H₂(g) + Br₂(g) $$ \to $$ 2HBr(g)

$$\Delta $$n_g = 2 - (1 + 1) = 0

So, $$\Delta $$H = $$\Delta $$E for this reaction.

The resonance energy provides extra stability to the benzene molecule so it has to over come for hydrogenation to take place.

So $$\Delta $$H = – 358.5 – (–150.4) = –208.1 kJ

$$\Delta $$G = $$\Delta $$H – T$$\Delta $$S

Now, at equilibrium $$\Delta $$G = 0

0 = $$\Delta $$H – T$$\Delta $$S

$$ \Rightarrow $$ 0 = 30000 –T (105)

$$ \Rightarrow $$ T = $${{30000} \over {105}}$$ = 285.7 K

$$\Delta $$G_system < 0, process is spontaneous.

$$\Delta $$G_system = 0, process is in equilibrium.

$$\Delta $$G_system > 0, process is not spontaneous.