Consider the following reversible reaction, A(g) + B(g) AB(g). The activation energy of the back

Q: Why is the equilibrium constant equal to the ratio of rate constants?

At equilibrium, the rate of the forward reaction exactly equals the rate of the backward reaction ($r_f = r_b$). Since $r_f = k_f[\text{Reactants}]$ and $r_b = k_b[\text{Products}]$, rearranging this gives $k_f/k_b = [\text{Products}]/[\text{Reactants}]$, which is the definition of the equilibrium constant $K_{eq}$.

Q: How did we know to use the Arrhenius equation here?

The problem explicitly mentions 'activation energy' and 'pre-exponential factor'. These are the hallmark variables of the Arrhenius equation ($k = A e^{-E_a/RT}$). Whenever you see these terms, it's a massive hint to express your rate constants in this form.

Q: I got confused expanding $\ln(4e^2)$. Can you explain that step again?

Absolutely! Logarithms turn multiplication into addition. So, $\ln(4 \cdot e^2)$ becomes $\ln(4) + \ln(e^2)$. We can write $4$ as $2^2$, so $\ln(2^2) = 2\ln(2)$. And the natural log of $e$ to any power is just that power, so $\ln(e^2) = 2$. Adding them up gives $2\ln(2) + 2$.

Q: Why did the $RT$ term cancel out in the exponent?

The exponent in the Arrhenius equation is $-E_a/RT$. When we divided $k_f$ by $k_b$, we subtracted their exponents: $(-E_{af}/RT) - (-E_{ab}/RT)$, which simplifies to $(E_{ab} - E_{af})/RT$. The problem stated that $E_{ab} - E_{af} = 2RT$. Substituting this in gives $2RT/RT$, and the $RT$ beautifully cancels out!

Q: Why did we take the absolute value at the end?

The calculated $\Delta G^\circ$ is $-8500 \text{ J/mol}$. The negative sign physically means the reaction is spontaneous. However, the question specifically asked for the 'absolute value' of $\Delta G^\circ$, which is just the magnitude without the sign. Always read the final sentence of a JEE question carefully to avoid losing marks on technicalities!

Question

Consider the following reversible reaction,
$	ext{A(g)} + 	ext{B(g)} ightleftharpoons 	ext{AB(g)}$.
The activation energy of the backward reaction exceeds that of the forward reaction by $2RT$ (in $	ext{J mol}^{-1}$).
If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of $\Delta G^	heta$ (in $	ext{J mol}^{-1}$) for the reaction at 300 K is____.
(Given ; $\ln (2) = 0.7$, $RT = 2500 	ext{ J mol}^{-1}$ at 300 K and G is the Gibbs energy)

Sigma Percentile · Accepted Answer

### The Beautiful Bridge Between Kinetics and Thermodynamics

Have you ever wondered how the microscopic dance of molecules—their collisions, their energy barriers, their frantic movements—translates into the grand, sweeping laws of thermodynamics? This problem is a perfect, elegant example of exactly that. We are going to build a bridge from the Arrhenius equation of chemical kinetics straight into the heart of Gibbs Free Energy.

Imagine you are standing at the foot of a mountain. To get to the valley on the other side, you have to climb over the peak. In chemistry, that peak is the **activation energy**.

### Analyzing the Setup

The problem gives us a reversible reaction:
$$ 	ext{A(g)} + 	ext{B(g)} ightleftharpoons 	ext{AB(g)} $$

We are told a crucial piece of information: the activation energy of the backward reaction ($E_{ab}$) exceeds that of the forward reaction ($E_{af}$) by $2RT$. Mathematically, this is:
$$ E_{ab} - E_{af} = 2RT $$

What does this mean physically? If it takes more energy to go backward than to go forward, the products must be at a lower energy state than the reactants. This is an exothermic reaction! The difference between these two activation energies is exactly the enthalpy change of the reaction, $\Delta H$.

### The Master Equation: Equilibrium Constant

Now, how do we connect these kinetic parameters to equilibrium? At equilibrium, the rate of the forward reaction equals the rate of the backward reaction. This fundamental balance gives us the equilibrium constant $K_{eq}$ as the ratio of the forward rate constant to the backward rate constant:
$$ K_{eq} = \frac{k_f}{k_b} $$

This is where the magic happens. We bring in the Arrhenius equation, $k = A e^{-E_a/RT}$, and substitute it for both rate constants:
$$ K_{eq} = \frac{A_f e^{-E_{af}/RT}}{A_b e^{-E_{ab}/RT}} $$

Let's group the terms to make it cleaner:
$$ K_{eq} = \left(\frac{A_f}{A_b}ight) e^{(E_{ab} - E_{af})/RT} $$

### Plugging in the Magic Numbers

The problem generously hands us the pieces of this puzzle. We know the pre-exponential factor of the forward reaction is 4 times that of the reverse, so $\frac{A_f}{A_b} = 4$. We also know the exponent numerator $E_{ab} - E_{af} = 2RT$. Let's substitute these in:
$$ K_{eq} = 4 \cdot e^{2RT/RT} $$
$$ K_{eq} = 4e^2 $$

Look at how beautifully the $RT$ terms cancelled out in the exponent! We now have a clean, exact value for our equilibrium constant.

### The Final Calculation: Gibbs Free Energy

We have arrived at the thermodynamic side of the bridge. The standard Gibbs Free Energy, $\Delta G^\circ$, is intimately tied to the equilibrium constant by the famous equation:
$$ \Delta G^\circ = -RT \ln K_{eq} $$

Let's substitute our $K_{eq}$:
$$ \Delta G^\circ = -RT \ln(4e^2) $$

Here is where many students make a silly mistake. Don't rush the logarithm! Use the properties of logs to expand it carefully: $\ln(ab) = \ln(a) + \ln(b)$.
$$ \ln(4e^2) = \ln(4) + \ln(e^2) $$
$$ \ln(4e^2) = 2\ln(2) + 2 $$

Now, we plug in the given values: $\ln(2) = 0.7$ and $RT = 2500 	ext{ J mol}^{-1}$.
$$ \ln(4e^2) = 2(0.7) + 2 = 1.4 + 2 = 3.4 $$

Finally, calculate $\Delta G^\circ$:
$$ \Delta G^\circ = -2500 	imes 3.4 = -8500 	ext{ J mol}^{-1} $$

The negative sign tells us the reaction is spontaneous in the forward direction under standard conditions. However, the question specifically asks for the **absolute value** of $\Delta G^\circ$.

Therefore, our final answer is **8500**.

Take a moment to appreciate what we just did. We took microscopic collision parameters and derived a macroscopic thermodynamic property. That is the true power of physical chemistry!

Enter Numerical Value:

Visualized Solution

$Energy Profile and Enthalpy$

$Equilibrium Constant K_{e q}$

$Evaluating K_{e q}$

$Standard Gibbs Free Energy$

$Final Calculation$

The Sigma Insight: Theories of Chemical Reaction

The Beautiful Bridge Between Kinetics and Thermodynamics

Analyzing the Setup

The Master Equation

Plugging in the Magic Numbers

The Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Profile and Enthalpy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equilibrium Constant Keq​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Keq​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Standard Gibbs Free Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation