Fig. 3.1 shows how the concentrations of substances A and B change with time when the reversible reaction A ⇌ B is carried out in a closed container at cons...

This question tests reading a concentration-time graph for the reversible reaction A ⇌ B and explaining equilibrium in terms of rates.

(a) Equilibrium is reached where both curves become horizontal (concentrations stop changing). On Fig. 3.1 this happens at 4 minutes [1].

(b) As the reaction proceeds the concentration of A falls, so the forward rate falls; the concentration of B rises, so the reverse rate rises [1]. When the falling forward rate meets the rising reverse rate the two become equal, so A is converted to B exactly as fast as B is converted back to A, and the concentrations no longer change [1].

(c) The forward reaction is fastest at the start, when the concentration of A is highest [1], and it slows down as A is used up, until it becomes constant at equilibrium [1]. A steep A curve at the start that levels off shows this.

(d) At equilibrium both the forward and reverse reactions continue at equal rates, so neither A nor B is ever completely used up; that is why both concentrations settle at fixed, non-zero values [1].

(e) Adding more A raises the concentration of a reactant, so the equilibrium shifts to the right, towards B [1]. The extra A speeds up the forward reaction, so more B is formed until the forward and reverse rates are once again equal and a new equilibrium is established [1].

Exam reminder: equilibrium on a concentration-time graph is where the lines go flat, not where they cross.