# Question #799b9

May 24, 2017

See below.

#### Explanation:

The forward and reverse reaction rates are different in the first 8 s, so the reaction is not at equilibrium in the first 8 s.

The forward and reverse reaction rates are identical after 8 s, so the reaction is at equilibrium after 8 s.

The upward sloping solid line represents the rate of the reverse reaction at any given time.

The rate of the reverse reaction would jump instantaneously to some higher value (say, $\text{2.0 mol·L"^"-1""s"^"-1}$) and then gradually decline to some equilibrium value (say, $\text{1.5 mol·L"^"-1""s"^"-1}$) over the next 8 s.

At the same time , the rate of the forward reaction would gradually increase to the same rate as the reverse reaction.