Animated Solution for Chemistry - Chemical Equilibrium: The standard Gibbs energy change at $300\text{ K}$ for the reaction, $2A\rightleftharpoons B+C$ is $2494.2\text{ J}$. At a given time, the composition of the reaction mixture is $[A]=\frac{1}{2},[B]=2$ and $[C]=\frac{1}{2}$. The reaction proceeds in the $[R=8.314{\text{ J K}}^{-1}{\text{ mol}}^{-1},e=2.718]$

Imagine you are standing on the side of a steep valley. If you drop a ball, it will naturally roll down to the lowest point, right? In chemistry, that lowest point is called Equilibrium, and the "height" of the valley is the Gibbs Free Energy ($G$). Every chemical reaction wants to reach the bottom of this valley where it is most stable.

In this problem, we are given a snapshot of a reaction: $2A\rightleftharpoons B+C$. We don't know if the ball is at the bottom, on the left slope, or on the right slope. To find out, we need two pieces of information: where we are right now (the Reaction Quotient, $Q$), and where the bottom of the valley is (the Equilibrium Constant, $K_c$).

The Reaction Quotient ($Q$)

The Reaction Quotient $Q$ is calculated exactly like the equilibrium constant, but using the current concentrations instead of equilibrium concentrations. It tells us the ratio of products to reactants at this exact moment.

Let's plug in the values given in the problem: $[A]=\frac{1}{2}$, $[B]=2$, and $[C]=\frac{1}{2}$.

So, our current "position" is $Q=4$. But is this the bottom of the valley? We need to find $K_c$ to know for sure.

The Equilibrium Constant ($K_c$)

We are given the standard Gibbs free energy change, $\Delta G^{\circ }=2494.2\text{ J}$. This value is the key to finding $K_c$ through the master equation of chemical thermodynamics:

Let's substitute the known values. The universal gas constant $R=8.314{\text{ J K}}^{-1}{\text{ mol}}^{-1}$ and the temperature $T=300\text{ K}$.

Notice the beautiful math here! If you multiply $8.314$ by $300$, you get exactly $2494.2$. This is a classic JEE setup designed to reward students who don't rush to their calculators.

Dividing both sides by $2494.2$, we get:

To solve for $K_c$, we take the exponential of both sides:

Which Way Will the Ball Roll?

Now we compare our current position ($Q$) with the equilibrium position ($K_c$).

We found that $Q=4$ and $K_c\approx 0.36$. Clearly, $Q>K_c$.

What does this mean physically? It means we have too many products compared to what the system wants at equilibrium. We are standing high up on the right side of the valley. To reach the stable bottom, the ball must roll backwards.

Therefore, the reaction will proceed in the reverse direction to consume the excess products and form more reactants until $Q$ drops down to match $K_c$.