Consider the reaction A B at 1000 K. At time t', the temperature of the system was increased to

Q: Why did we use $K_p$ instead of $K_c$?

The problem provides data in terms of partial pressures (bar), which naturally leads to calculating $K_p$. Furthermore, standard Gibbs free energy for gases is defined using standard states of 1 bar, making $K_p$ the correct constant to use in $\Delta G^\circ = -RT \ln K_p$.

Q: Why is the partial pressure of A constant at 1 bar? Doesn't it react?

The problem explicitly states 'the partial pressure of A was maintained at 1 bar'. This implies the experimental setup is not a simple closed vessel; A is being continuously supplied or removed to keep its pressure exactly at 1 bar, acting like a large reservoir.

Q: What is the significance of the spike to 115 bar in the graph?

The spike represents a transient non-equilibrium state immediately after the temperature is doubled. The sudden increase in temperature causes a rapid expansion/pressure increase, followed by the chemical equilibrium shifting to its new steady state at 100 bar. For our thermodynamic calculations, only the final flat equilibrium lines matter.

Q: Can I use $\log_{10}$ instead of natural log ($\ln$)?

Yes, you can! Since $\ln(x) = 2.303 \log_{10}(x)$, the factor of 2.303 will appear in both the numerator and denominator and cancel out perfectly. The ratio will remain exactly the same.

Q: Why is the ratio 0.25 and not 4?

You must be careful with the order of the ratio requested. The question asks for the ratio of $\Delta G^\circ$ at 1000 K to that at 2000 K. So it is $\frac{\Delta G^\circ_{1000}}{\Delta G^\circ_{2000}}$. If you flip it, you get 4, which is a classic trap!

Question

Consider the reaction $	ext{A} ightleftharpoons 	ext{B}$ at $1000	ext{ K}$. At time $t'$, the temperature of the system was increased to $2000	ext{ K}$ and the system was allowed to reach equilibrium. Throughout this experiment the partial pressure of A was maintained at $1	ext{ bar}$. Given below is the plot of the partial pressure of B with time. What is the ratio of the standard Gibbs energy of the reaction at $1000	ext{ K}$ to that at $2000	ext{ K}$?

Sigma Percentile · Accepted Answer

Imagine you are a scientist observing a chemical reaction in a highly controlled, closed chamber. You are monitoring the reaction $	ext{A} ightleftharpoons 	ext{B}$. The pressure of reactant A is artificially maintained at exactly $1	ext{ bar}$ throughout the entire experiment—think of it as being connected to a massive reservoir of gas A. Your only job is to watch the pressure of product B evolve over time as you manipulate the temperature.

This is exactly the scenario presented in this beautiful JEE Advanced problem. It elegantly combines graphical interpretation with the core principles of chemical thermodynamics. Let's break down the journey step-by-step.

### Decoding the Graph: Extracting the Equilibrium Pressures

The graph provided is the heart of the problem. It plots the partial pressure of B against time. Notice how the curve features two distinct, perfectly horizontal flat lines. In the world of chemical kinetics and thermodynamics, a flat line on a concentration or pressure vs. time graph screams one word: **Equilibrium**.

Before time $t'$, the system is held at a temperature of $1000	ext{ K}$. The graph shows that the pressure of B is perfectly stable at $10	ext{ bar}$. This is our first crucial data point: at $1000	ext{ K}$, the equilibrium partial pressure of B is $P_B = 10	ext{ bar}$.

Suddenly, at time $t'$, the temperature is doubled to $2000	ext{ K}$. The system experiences a thermal shock! The pressure spikes dramatically (a consequence of the sudden heating and rapid kinetic shift), but eventually, the system finds a new balance. It settles into a new horizontal line at $100	ext{ bar}$. This gives us our second data point: at $2000	ext{ K}$, the new equilibrium partial pressure of B is $P_B = 100	ext{ bar}$.

### The Equilibrium Constant: The Bridge to Thermodynamics

To connect these physical observations to thermodynamic properties, we need the equilibrium constant, $K_p$. For our simple reaction $	ext{A} ightleftharpoons 	ext{B}$, the equilibrium constant is defined as the ratio of the partial pressures of the products to the reactants:

$$K_p = \frac{P_B}{P_A}$$

The problem explicitly states a massive constraint: $P_A$ is maintained at $1	ext{ bar}$ throughout the experiment. This makes our calculation incredibly straightforward. The equilibrium constant $K_p$ is numerically equal to the partial pressure of B!

Let's calculate $K_p$ for both temperatures:
At $1000	ext{ K}$: 
$$K_{1000} = \frac{10}{1} = 10$$

At $2000	ext{ K}$: 
$$K_{2000} = \frac{100}{1} = 100$$

### The Master Equation: Gibbs Free Energy

Now we bring in the heavy machinery of thermodynamics. The standard Gibbs free energy change, $\Delta G^\circ$, is fundamentally linked to the equilibrium constant by the master equation:

$$\Delta G^\circ = -RT \ln K_p$$

This equation tells us how the intrinsic thermodynamic drive of the reaction ($\Delta G^\circ$) dictates the final equilibrium position ($K_p$) at a given temperature ($T$).

We need to find the ratio of the standard Gibbs energy at $1000	ext{ K}$ to that at $2000	ext{ K}$. Let's set up the fraction:

$$	ext{Ratio} = \frac{\Delta G^\circ_{1000}}{\Delta G^\circ_{2000}} = \frac{-R(1000) \ln(K_{1000})}{-R(2000) \ln(K_{2000})}$$

### The Final Ratio: Mathematical Elegance

First, let's clear the clutter. The negative signs and the universal gas constant $R$ cancel out immediately from the numerator and denominator.

$$	ext{Ratio} = \frac{1000 \ln(10)}{2000 \ln(100)}$$

Here is where a solid grasp of logarithm properties saves the day. Do not leave $\ln(100)$ as it is. Recognize that $100$ is simply $10^2$. Using the power rule of logarithms, $\ln(x^n) = n \ln(x)$, we can rewrite the denominator:

$$\ln(100) = \ln(10^2) = 2 \ln(10)$$

Substitute this beautiful simplification back into our ratio:

$$	ext{Ratio} = \frac{1000 \ln(10)}{2000 	imes 2 \ln(10)}$$

Look at that elegance! The $\ln(10)$ terms cancel out completely. We are left with pure, simple arithmetic:

$$	ext{Ratio} = \frac{1000}{4000} = \frac{1}{4} = 0.25$$

**The final answer is 0.25.**

### Beyond the Problem: Le Chatelier's Whisper

Before we close the book on this problem, let's think like an examiner. What hidden story does this data tell?

Notice that as we increased the temperature from $1000	ext{ K}$ to $2000	ext{ K}$, our equilibrium constant $K_p$ increased from $10$ to $100$. The reaction shifted forward to produce more B at the higher temperature. According to **Le Chatelier's Principle**, a system will shift to absorb added heat if the reaction is endothermic. Therefore, because heating the system drove the reaction forward, we can definitively conclude that the forward reaction $	ext{A} ightarrow 	ext{B}$ is an **endothermic process** ($\Delta H > 0$).

Always look for these deeper physical insights—they are what separate good students from great physicists and chemists!

Enter Numerical Value:

Visualized Solution

$Analyzing the Graph$

$Equilibrium at 1000 K$

$Equilibrium at 2000 K$

$Equilibrium Constant K_{p}$

$Calculating K_{1000} and K_{2000}$

$Standard Gibbs Free Energy$

$Setting up the Ratio$

$Substituting Values$

$Simplifying Logarithms$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Entropy and Free Energy

Decoding the Graph

The Equilibrium Constant

The Master Equation

The Final Ratio

Beyond the Problem

Similar Questions

The standard reaction Gibbs energy for a chemical reaction at an absolute temperature $T$ is given by, $Δ_{r} G^{\circ} = A - B T$ Where $A$ and $B$ are non-zero constants. Which of the following is true about this reaction?

For the chemical reaction, $X ⇌ Y$ , the standard reaction Gibbs energy depends on temperature $T$ (in K) as $Δ_{r} G^{\circ} (in kJ mol^{- 1}) = 120 - \frac{3}{8} T$ The major component of the reaction mixture at $T$ is

For the reaction; $A (l) ⟶ 2 B (g)$ $Δ U = 2.1 kcal, Δ S = 20 cal K^{- 1}$ at $300 K$ . Hence, $Δ G$ in kcal is ......

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For a given chemical reaction, $A ⟶ B$ at $300 K$ the free energy change is $- 49.4 kJ mol^{- 1}$ and the enthalpy of reaction is $51.4 kJ mol^{- 1}$ . The entropy change of the reaction is ...... $JK^{- 1} mol^{- 1}$ .

For a particular reversible reaction, at temperature $T$ , $Δ H$ and $Δ S$ were found to be both +ve. If $T_{e}$ is the temperature at equilibrium, the reaction would be spontaneous when

Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equilibrium at 1000 K

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equilibrium at 2000 K

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating K1000​ and K2000​

.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; } Setting up the Ratio

.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; } Simplifying Logarithms

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

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