Bromine and hydrogen react in the gas phase to form hydrogen bromide: \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) .\) The reaction enthalpy is \(\Delta H^{\circ}=-6 \mathrm{~kJ} .\) (a) To increase the equilibrium yield of hydrogen bromide would you use high or low temperature? (b) Could you increase the equilibrium yield of hydrogen bromide by controlling the pressure of this reaction? If so, would high or low pressure favor formation of \(\mathrm{HBr}(g) ?\)

Since the reaction is exothermic (\(\Delta H^\circ < 0\)), using Le Chatelier's principle, we know that increasing the temperature will shift the reaction towards the side with more energy (the reactants) and decreasing the temperature will shift the reaction towards the side with less energy (the products). Therefore, to increase the equilibrium yield of hydrogen bromide, we should use a low temperature.

To examine the effect of pressure on the equilibrium yield of HBr, we need to compare the number of moles of reactant gases (H2 and Br2) with the number of moles of product gas (HBr). In this reaction, we have: \[\mathrm{H}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2 \mathrm{HBr}(g)\] There are initially 1 mole of H2 and 1 mole of Br2, making a total of 2 moles of reactant gases. After the reaction, there are 2 moles of HBr gas. Since the number of moles of gas on both sides of the reaction is equal (2 moles in both cases), changing the pressure will not significantly affect the position of the equilibrium. Thus, controlling the pressure of this reaction is not a suitable way to increase the equilibrium yield of hydrogen bromide.