For the equilibrium , the variation of the rate of the forward (a) and reverse (b) reaction with time is given by:

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At the start of the reaction, the forward rate is maximum (high [A]) and the reverse rate is zero (no B present). As the reaction proceeds, [A] decreases so the forward rate decreases, while [B] increases so the reverse rate increases. At equilibrium, both rates become equal and constant.

The correct graph shows the forward rate (a) starting high and decreasing, the reverse rate (b) starting at zero and increasing, and both meeting at a constant value at equilibrium.

Answer: Option (2)

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