Given,

_{} for the reaction i.e., H_{2(g)} + Br_{2(g)} ↔ 2HBr_{(g}_{)} is 1.6x10^{5}^{ }Therefore, for the reaction 2HBr_{(g}) ↔ H_{2(g) }+ Br_{2(g)} the equilibrium constant will be,

Now, let p be the pressure of both H_{2 }and Br_{2 }at equilibrium.

Now, we can write,

Therefore, at equilibrium,