When oxygen gas is reacted with hydrogen bromide, bromine gas is produced and heat is evolved according to the equation:

4HBr(g) + O2(g)
 Br2(g) + 2H2O(g)
ΔH = –155 kJ mol–1
Predict the effect on the equilibrium position and the equilibrium constant of the following changes.
a Extra HBr is added to the container.
b The vessel is heated.
c Some unreactive argon gas is pumped into the container.
d The volume of the reaction vessel is decreased.
e A catalyst is employed.

1 answer

a) If extra HBr is added to the container, the equilibrium position will shift to the right, favoring the formation of more products (Br2 and H2O). This is because according to Le Chatelier's principle, when more reactant (HBr) is added, the equilibrium will shift in the direction that consumes the added reactant. The equilibrium constant, however, will remain unchanged.

b) If the vessel is heated, the equilibrium position will shift to the left, favoring the formation of more reactants (HBr and O2). This is because the reaction is exothermic (heat is evolved), and according to Le Chatelier's principle, increasing the temperature will favor the endothermic direction to consume the excess heat. The equilibrium constant will not be affected.

c) If unreactive argon gas is pumped into the container, it will have no effect on the equilibrium position or the equilibrium constant. This is because argon is unreactive and does not participate in the chemical reaction.

d) If the volume of the reaction vessel is decreased, the equilibrium position will shift to the side with fewer moles of gas. In this case, it will shift to the right, favoring the formation of more products (Br2 and H2O). This is because according to Le Chatelier's principle, when the volume is decreased, the equilibrium will shift in the direction that reduces the number of moles of gas. The equilibrium constant will not be affected.

e) If a catalyst is employed, it will increase the rate of the forward and reverse reactions equally, but it will not affect the equilibrium position or the equilibrium constant. This is because a catalyst provides an alternate pathway for the reaction, reducing the activation energy for both the forward and reverse reactions. The equilibrium position is determined by the balance of forward and reverse reactions, which is not affected by the presence of a catalyst. The equilibrium constant will remain unchanged.