The plot of log k_f versus 1/T for a reversible reaction A (g) ⇌ P (g) is shown. [JEE(Advanced) 2023

Q: Why is the Arrhenius plot a straight line?

Because taking the logarithm of the Arrhenius equation $k = A e^{-E_a/RT}$ gives $\log k = \log A - \frac{E_a}{2.303RT}$, which is in the form $y = mx + c$.

Q: How do we relate the equilibrium constant to rate constants?

At equilibrium, the rate of the forward reaction equals the rate of the backward reaction, which leads to $K = \frac{k_f}{k_b}$.

Q: Why did we subtract the activation energies?

When dividing the rate constants $k_f$ and $k_b$, their exponential terms divide, which means their exponents subtract: $e^{-E_f/RT} / e^{-E_b/RT} = e^{-(E_f - E_b)/RT}$.

Q: What is the significance of the pre-exponential factor?

It represents the frequency of collisions between reactant molecules with the correct orientation, even if they don't all have enough energy to react.

Q: Why take the absolute value of $\log k_b$?

The question specifically asks for $|\log k_b|$, which is the magnitude. Since $\log k_b$ is $-5$, its absolute value is $5$.

Question

The plot of $\log k_f$ versus $\frac{1}{T}$ for a reversible reaction $A (g) ightleftharpoons P (g)$ is shown. [JEE(Advanced) 2023]
Pre-exponential factors for the forward and backward reactions are $10^{15} 	ext{ s}^{-1}$ and $10^{11} 	ext{ s}^{-1}$, respectively. If the value of $\log K$ for the reaction at $500 	ext{ K}$ is $6$, the value of $|\log k_b|$ at $250 	ext{ K}$ is ______.
[$K = 	ext{equilibrium constant of the reaction}$
$k_f = 	ext{rate constant of forward reaction}$
$k_b = 	ext{rate constant of backward reaction}$]

Sigma Percentile · Accepted Answer

The journey of mastering Chemical Kinetics often brings us to the beautiful intersection of thermodynamics and reaction rates. This problem from JEE Advanced 2023 is a perfect example of how multiple concepts weave together to form a comprehensive challenge. It tests your ability to read an Arrhenius plot, manipulate logarithmic equations, and connect the equilibrium constant with forward and backward rate constants.

Let's break down this elegant problem step-by-step and uncover the physics and math hidden within the equations.

### Decoding the Arrhenius Plot

The problem begins with a visual anchor: a plot of $\log k_f$ versus $\frac{1}{T}$. This is the classic Arrhenius plot.

The Arrhenius equation for the forward reaction is given by:
$$k_f = A_f e^{-\frac{E_f}{RT}}$$

To match the linear graph, we take the base-10 logarithm on both sides:
$$\log k_f = \log A_f - \frac{E_f}{2.303 RT}$$

This equation is now in the form of a straight line, $y = mx + c$, where our y-axis is $\log k_f$ and our x-axis is $\frac{1}{T}$. The slope of this line is $-\frac{E_f}{2.303 R}$, and the y-intercept is $\log A_f$.

The graph provides us with a specific coordinate point: when $\frac{1}{T} = 0.002 	ext{ K}^{-1}$, the value of $\log k_f$ is $9$.

We are also given the pre-exponential factor for the forward reaction, $A_f = 10^{15} 	ext{ s}^{-1}$.

Let's substitute these known values into our linear equation:
$$9 = \log(10^{15}) - \frac{E_f}{2.303 R} (0.002)$$

Since $\log(10^{15}) = 15$, the equation simplifies to:
$$9 = 15 - \frac{E_f}{2.303 R} (0.002)$$

Rearranging the terms to isolate the activation energy component:
$$\frac{E_f}{2.303 R} (0.002) = 15 - 9 = 6$$

$$\frac{E_f}{2.303 R} = \frac{6}{0.002} = 3000$$

We have successfully extracted a crucial piece of information from the graph. We now know the value of $\frac{E_f}{2.303 R}$, which will be instrumental in the next phase of the problem.

### Bridging Kinetics and Equilibrium

The problem introduces the equilibrium constant, $K$. A fundamental principle of chemical kinetics is that at equilibrium, the rate of the forward reaction equals the rate of the backward reaction. This leads to the relationship:
$$K = \frac{k_f}{k_b}$$

Taking the base-10 logarithm on both sides gives us:
$$\log K = \log k_f - \log k_b$$

We can expand $\log k_f$ and $\log k_b$ using their respective Arrhenius equations:
$$\log K = \left( \log A_f - \frac{E_f}{2.303 RT} ight) - \left( \log A_b - \frac{E_b}{2.303 RT} ight)$$

Grouping the pre-exponential factors and the activation energy terms together, we get a master equation:
$$\log K = \log\left(\frac{A_f}{A_b}ight) - \frac{E_f - E_b}{2.303 RT}$$

This equation beautifully connects the thermodynamic equilibrium constant with the kinetic parameters of both the forward and backward reactions.

### Solving for the Backward Reaction

The problem states that at a temperature of $T = 500 	ext{ K}$, the value of $\log K$ is $6$. We are also given the pre-exponential factor for the backward reaction, $A_b = 10^{11} 	ext{ s}^{-1}$.

Let's substitute these values into our master equation:
$$6 = \log\left(\frac{10^{15}}{10^{11}}ight) - \frac{E_f - E_b}{2.303 R (500)}$$

Simplifying the logarithmic term:
$$6 = \log(10^4) - \frac{E_f - E_b}{2.303 R (500)}$$

$$6 = 4 - \frac{E_f - E_b}{2.303 R (500)}$$

Rearranging to solve for the activation energy difference:
$$\frac{E_f - E_b}{2.303 R (500)} = 4 - 6 = -2$$

Multiplying both sides by $500$:
$$\frac{E_f - E_b}{2.303 R} = -1000$$

Or, equivalently:
$$\frac{E_b - E_f}{2.303 R} = 1000$$

We already found that $\frac{E_f}{2.303 R} = 3000$. We can substitute this into our new equation to find the backward activation energy term:
$$\frac{E_b}{2.303 R} - 3000 = 1000$$

$$\frac{E_b}{2.303 R} = 4000$$

We now have all the pieces of the puzzle for the backward reaction.

### The Final Calculation

The ultimate goal is to find the value of $|\log k_b|$ at a new temperature, $T = 250 	ext{ K}$.

We write the Arrhenius equation for the backward reaction:
$$\log k_b = \log A_b - \frac{E_b}{2.303 RT}$$

Substitute the known values: $A_b = 10^{11}$, $T = 250$, and $\frac{E_b}{2.303 R} = 4000$:
$$\log k_b = \log(10^{11}) - \frac{4000}{250}$$

$$\log k_b = 11 - 16$$

$$\log k_b = -5$$

The question specifically asks for the absolute value, $|\log k_b|$:
$$|\log k_b| = |-5| = 5$$

**The final answer is 5.**

This problem is a fantastic exercise in algebraic manipulation and conceptual clarity. By systematically breaking down the given information and applying fundamental kinetic principles, we navigated through the complexities to arrive at a clean, elegant solution. Always remember to trust the equations and let the math guide you!

Enter Numerical Value:

Visualized Solution

$Arrhenius Equation$

$Extracting Data from Graph$

$Substituting Values$

$Solving for Activation Energy Term$

$Equilibrium Constant Relation$

$Substituting Equilibrium Data$

$Solving for Backward Activation Energy Term$

$Calculating E_{b} Term$

$Backward Reaction at 250 K$

$Final Answer$

The Sigma Insight: Theories of Chemical Reaction

Decoding the Arrhenius Plot

Bridging Kinetics and Equilibrium

Solving for the Backward Reaction

The Final Calculation

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For following reactions : $A 700 K Product$ ; $A 500 K Catalyst Product$ It was found that the $E_{a}$ is decreased by $30 kJ/mol$ in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):

Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Extracting Data from Graph

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Activation Energy Term

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Backward Activation Energy Term

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Backward Reaction at 250 K

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