Animated Solution for Chemistry - Chemical Equilibrium: For the equilibrium $A\rightleftharpoons B$, the variation of the rate of the forward (a) and reverse (b) reaction with time is given by

Let's dive into the fascinating world of chemical equilibrium. Imagine a reversible reaction where reactant $A$ turns into product $B$. This is not a one-way street; it is a dynamic two-way process. We have two distinct reactions happening simultaneously within the same vessel: the forward reaction ($A\to B$) and the reverse reaction ($B\to A$).

To truly understand what equilibrium means, we must track the rates of these two reactions over time. The rate of a reaction is fundamentally dependent on the concentration of its reactants. Let's set up a mental graph to visualize this journey.

Starting Strong

At the very beginning, when time $t=0$, our reaction vessel contains only reactant $A$. There is absolutely no product $B$ present yet. Because the concentration of $A$, denoted as $[A]$, is at its absolute maximum, the forward reaction rate ($r_f$) is also at its peak.

However, as the clock ticks, $A$ gets consumed to form $B$. Its concentration steadily drops. According to the Law of Mass Action, as $[A]$ decreases, the rate of the forward reaction $r_f$ must also gradually decrease over time. If we plot this on a graph, the curve for the forward rate starts high on the y-axis and slopes downwards, representing a continuous deceleration.

Building Momentum

Now, what about the reverse reaction? Initially, since there is no product $B$ at all ($[B]=0$), the reverse reaction cannot even begin. Its rate ($r_b$) is exactly zero.

But the forward reaction is busy producing $B$. As more and more $B$ accumulates in the vessel, the reverse reaction starts to pick up speed. The concentration $[B]$ is increasing, and therefore, the rate of the reverse reaction $r_b$ increases steadily. On our graph, the curve for the reverse rate starts from the origin $(0,0)$ and climbs upwards, gaining momentum as time goes on.

Dynamic Equilibrium

So, we have a situation where the forward rate is continuously dropping and the reverse rate is continuously rising. They are on a collision course. Eventually, there comes a magical moment in time where these two rates perfectly match each other.

This specific point in time is what we call $t_{eq}$, the time equilibrium is achieved. This is the state of dynamic equilibrium! The rate at which $A$ turns into $B$ is exactly equal to the rate at which $B$ turns back into $A$.

Because the rates are equal, for every molecule of $A$ that disappears, another molecule of $A$ is instantly created by the reverse reaction. Consequently, the macroscopic concentrations of $A$ and $B$ stop changing. They become constant. On our graph, because the rates are now identical and unchanging, the two curves merge into a single, flat, horizontal line. The system has stabilized, even though at the microscopic level, the reactions are still furiously happening.

Comparing our logical deduction with the given options makes the answer crystal clear.

Graph (a) is the only one that correctly shows the forward rate starting high and decreasing, the reverse rate starting at zero and increasing, and most importantly, the two curves meeting and merging into a single horizontal line. This perfectly illustrates the condition $r_f=r_b$.

The other graphs show the rates leveling off at different values (where $r_f eq r_b$), which contradicts the very definition of chemical equilibrium. If the rates were different, the concentrations would continue to change, and the system would not be at equilibrium. Therefore, option (a) is the undeniably correct representation.