In a one-litre flask, 6 moles of A undergoes the reaction A (g) P (g). The progress of product f

Q: Why did we take the volume as 1 L?

The problem explicitly states 'In a one-litre flask'. This makes the number of moles numerically equal to the molar concentration, simplifying our calculations immensely.

Q: Why is $K_p$ equal to $K_c$ here?

For the reaction $A(g) \rightleftharpoons P(g)$, the change in the number of moles of gas ($\Delta n_g$) is $1 - 1 = 0$. Since $K_p = K_c(RT)^{\Delta n_g}$, when $\Delta n_g = 0$, $K_p$ is exactly equal to $K_c$.

Q: How do we know the reaction has reached equilibrium from the graph?

The concentration of product P becomes constant over time, which is indicated by the curves becoming horizontal (asymptotic). At equilibrium, the macroscopic properties like concentration do not change.

Q: Why is $\ln(1/2)$ equal to $-\ln 2$?

According to the properties of logarithms, $\ln(a/b) = \ln a - \ln b$. So, $\ln(1/2) = \ln 1 - \ln 2$. Since $\ln 1 = 0$, it simplifies perfectly to $-\ln 2$.

Q: Can we use $\log_{10}$ instead of $\ln$?

The standard thermodynamic relation is $\Delta G^\ominus = -RT \ln K$, which uses the natural logarithm (base $e$). The problem also asks for the answer in terms of $\ln x$, so we must stick to natural logarithms to get the correct value of $x$.

Question

In a one-litre flask, 6 moles of A undergoes the reaction A (g) $\rightleftharpoons$ P (g). The progress of product formation at two temperatures (in Kelvin), $T_1$ and $T_2$, is shown in the figure:

If $T_1 = 2T_2$ and $(\Delta G_2^\ominus - \Delta G_1^\ominus) = RT_2 \ln x$, then the value of x is ______.

[$\Delta G_1^\ominus$ and $\Delta G_2^\ominus$ are standard Gibb's free energy change for the reaction at temperatures $T_1$ and $T_2$, respectively.]

Sigma Percentile · Accepted Answer

Welcome to the beautiful intersection of chemical kinetics, equilibrium, and thermodynamics! This problem is a masterpiece from JEE Advanced that tests your ability to read a graph, set up an equilibrium state, and seamlessly transition into the abstract world of Gibbs Free Energy. Let's break it down step by step.

### Decoding the Graph: The Story of Equilibrium

The graph provided is a visual story of a chemical reaction approaching its final destination: equilibrium. The x-axis represents time, and the y-axis represents the concentration of our product, P.

Notice how the curves rise steeply at first but eventually flatten out into horizontal lines? That flattening is the hallmark of equilibrium. It means the forward and backward reaction rates have become equal, and the macroscopic concentration of P has stopped changing.

By simply reading the asymptotes of the graph, we can extract crucial data:
- At temperature $T_1$, the equilibrium concentration is $[P]_{	ext{eq}} = 4 	ext{ M}$.
- At temperature $T_2$, the equilibrium concentration is $[P]_{	ext{eq}} = 2 	ext{ M}$.

### The ICE Table: Uncovering the Hidden Reactant

We know the equilibrium concentration of the product, but to find the equilibrium constant $K$, we also need the equilibrium concentration of the reactant, A.

The reaction is a simple 1:1 stoichiometry: $A(g) ightleftharpoons P(g)$.

We are told the reaction happens in a 1-litre flask with an initial 6 moles of A. This makes our initial concentration $[A]_0 = 6 	ext{ M}$. Because 1 mole of A reacts to form 1 mole of P, the amount of A that disappears is exactly equal to the amount of P that appears.

Therefore, at equilibrium, the concentration of A is simply the initial amount minus what reacted: 
$$[A]_{	ext{eq}} = 6 - [P]_{	ext{eq}}$$

### Calculating the Equilibrium Constants

Now, let's calculate the equilibrium constants for both temperatures. The equilibrium constant $K$ is the ratio of the product concentration to the reactant concentration.

**For Temperature $T_1$:**
- $[P]_{	ext{eq}} = 4 	ext{ M}$
- $[A]_{	ext{eq}} = 6 - 4 = 2 	ext{ M}$
- $K_1 = \frac{4}{2} = 2$

**For Temperature $T_2$:**
- $[P]_{	ext{eq}} = 2 	ext{ M}$
- $[A]_{	ext{eq}} = 6 - 2 = 4 	ext{ M}$
- $K_2 = \frac{2}{4} = \frac{1}{2}$

### The Master Equation: Enter Gibbs Free Energy

How do we connect these macroscopic equilibrium constants to the fundamental thermodynamic driving force of the reaction? We use the legendary master equation:

$$\Delta G^\ominus = -RT \ln K$$

Let's plug in our calculated $K$ values for both temperatures:
- $\Delta G_1^\ominus = -RT_1 \ln 2$
- $\Delta G_2^\ominus = -RT_2 \ln \left(\frac{1}{2}ight)$

Remember a handy property of logarithms: $\ln(1/x) = -\ln x$. Therefore, $\ln(1/2) = -\ln 2$. This transforms our second equation into:
- $\Delta G_2^\ominus = RT_2 \ln 2$

### The Final Algebraic Symphony

We are given a crucial constraint in the problem: $T_1 = 2T_2$. Let's substitute this into our expression for $\Delta G_1^\ominus$ to get everything in terms of $T_2$:

$$\Delta G_1^\ominus = -R(2T_2) \ln 2 = -2RT_2 \ln 2$$

The question asks us to evaluate the expression $(\Delta G_2^\ominus - \Delta G_1^\ominus)$. Let's perform the subtraction carefully:

$$\Delta G_2^\ominus - \Delta G_1^\ominus = RT_2 \ln 2 - (-2RT_2 \ln 2)$$
$$\Delta G_2^\ominus - \Delta G_1^\ominus = 3RT_2 \ln 2$$

To match the format given in the question ($RT_2 \ln x$), we need to move the coefficient 3 inside the logarithm. Using the power rule of logarithms ($a \ln b = \ln(b^a)$), we get:

$$3RT_2 \ln 2 = RT_2 \ln(2^3) = RT_2 \ln 8$$

Comparing this magnificent result to the given expression $RT_2 \ln x$, it is crystal clear that **$x = 8$**.

Enter Numerical Value:

Visualized Solution

$[P]_{eq} at T_{1}, T_{2}$

$A (g) ⇌ P (g)$

$K_{1} at T_{1}$

$K_{2} at T_{2}$

$Δ G^{⊖} = - R T ln K$

$Δ G_{1}^{⊖}, Δ G_{2}^{⊖}$

$T_{1} = 2 T_{2}$

$Δ G_{2}^{⊖} - Δ G_{1}^{⊖}$

$x = 8$

The Sigma Insight: Entropy and Free Energy

Decoding the Graph

The ICE Table

Calculating the Equilibrium Constants

The Master Equation

The Final Algebraic Symphony

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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; } [P]eq​ at T1​,T2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A(g)⇌P(g)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K1​ at T1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K2​ at T2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔG⊖=−RTlnK

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔG1⊖​,ΔG2⊖​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​=2T2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔG2⊖​−ΔG1⊖​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=8