For the reaction, A(g) ⇌ B(g) at 495 K, ΔrG° = -9.478 kJ mol⁻¹. If we start the reaction in a closed

Q: Why do we multiply $\Delta_r G^\circ$ by 1000?

The universal gas constant $R$ is given in Joules ($8.314\text{ J mol}^{-1}\text{ K}^{-1}$), but $\Delta_r G^\circ$ is provided in kiloJoules ($-9.478\text{ kJ/mol}$). We must multiply by 1000 to ensure unit consistency across the equation.

Q: Why did the volume $V$ disappear from the equilibrium expression?

Since the reaction is $A(g) \rightleftharpoons B(g)$, the number of moles of gas is the same on both sides ($\Delta n_g = 0$). When you write $K = \frac{[B]}{[A]} = \frac{n_B/V}{n_A/V}$, the volume terms in the numerator and denominator cancel out perfectly.

Q: How did we calculate the complex denominator without a calculator?

In JEE problems, the numbers are often rigged to cancel out elegantly. If you notice, the product of $2.303 \times 8.314 \times 495$ is approximately $9478$, which perfectly matches the numerator, simplifying the entire fraction to $1$.

Q: What does a negative $\Delta_r G^\circ$ physically mean here?

A negative standard Gibbs free energy change means that the forward reaction is thermodynamically favorable under standard conditions. This is why our calculated equilibrium constant $K$ is greater than $1$ ($K=10$), indicating that products are favored at equilibrium.

Q: Can we use millimoles directly in the ICE table instead of converting to moles?

Yes! As long as you are consistent. Since the question asks for the final answer in millimoles, keeping the initial amount as $22\text{ millimoles}$ saves us from unnecessary conversions and potential decimal errors.

Question

For the reaction, $A(g) ightleftharpoons B(g)$ at $495	ext{ K}$, $\Delta_r G^\circ = -9.478	ext{ kJ mol}^{-1}$.
If we start the reaction in a closed container at $495	ext{ K}$ with $22$ millimoles of $A$, the amount of $B$ in the equilibrium mixture is ......... millimoles (Round off to the nearest integer).
$[R = 8.314	ext{ J mol}^{-1}	ext{ K}^{-1}, \ln 10 = 2.303]$

Sigma Percentile · Accepted Answer

This problem is a beautiful intersection of Chemical Thermodynamics and Chemical Equilibrium. It tests your ability to bridge the macroscopic energy changes of a reaction with the microscopic distribution of molecules at equilibrium. Let's break down the journey step-by-step.

### The Thermodynamic Bridge

We are given a simple gaseous reaction: $A(g) ightleftharpoons B(g)$ occurring in a closed container at a constant temperature of $495	ext{ K}$. The key piece of information provided is the standard Gibbs free energy change, $\Delta_r G^\circ = -9.478	ext{ kJ mol}^{-1}$.

How does this energy value help us find the amounts of gases? The bridge between thermodynamics and equilibrium is the master equation:

$$\Delta_r G^\circ = -2.303 RT \log_{10} K$$

This equation tells us that the standard free energy change dictates the position of equilibrium. A negative $\Delta_r G^\circ$ implies that the forward reaction is thermodynamically favorable, meaning we expect the equilibrium constant $K$ to be greater than $1$.

### Decoding the Math

Let's substitute our known values into the master equation. **Crucially, we must ensure unit consistency.** Since the universal gas constant $R$ is given in $	ext{J mol}^{-1}	ext{ K}^{-1}$, we must convert $\Delta_r G^\circ$ from kilojoules to joules by multiplying by $1000$.

$$-9.478 	imes 10^3 = -2.303 	imes 8.314 	imes 495 	imes \log_{10} K$$

Isolating $\log_{10} K$, the negative signs on both sides cancel out:

$$\log_{10} K = \frac{9478}{2.303 	imes 8.314 	imes 495}$$

At first glance, the denominator looks like a nightmare to calculate without a calculator. However, in JEE problems, the numbers are often meticulously crafted. If you estimate or carefully multiply the denominator, you will find that $2.303 	imes 8.314 	imes 495 \approx 9478$.

This beautiful cancellation simplifies our fraction to exactly $1$:

$$\log_{10} K = 1 \implies K = 10^1 = 10$$

### The ICE Table

Now that we have our equilibrium constant $K = 10$, we can determine the actual composition of the mixture. We set up an ICE (Initial, Change, Equilibrium) table. We start with $22	ext{ mmol}$ of $A$ and $0	ext{ mmol}$ of $B$.

Let $x$ be the number of millimoles of $A$ that react to reach equilibrium. Because the stoichiometry is $1:1$, exactly $x$ millimoles of $B$ will be produced.

*   **Initial:** $n_A = 22$, $n_B = 0$
*   **Change:** $\Delta n_A = -x$, $\Delta n_B = +x$
*   **Equilibrium:** $n_A = 22 - x$, $n_B = x$

The equilibrium constant $K$ is defined as the ratio of the concentrations of the products to the reactants. Since both gases share the same container volume $V$, the volume terms cancel out perfectly:

$$K = \frac{[B]}{[A]} = \frac{n_B / V}{n_A / V} = \frac{n_B}{n_A}$$

### The Final Calculation

Substituting our equilibrium expressions into the $K$ equation gives us a simple linear equation:

$$10 = \frac{x}{22 - x}$$

Cross-multiplying to solve for $x$:

$$220 - 10x = x$$

$$11x = 220$$

$$x = 20$$

Since $x$ represents the number of millimoles of $B$ at equilibrium, we have our final answer. The equilibrium mixture contains exactly **$20	ext{ millimoles}$** of gas $B$.

Enter Numerical Value:

Visualized Solution

$A (g) ⇌ B (g)$

$Δ_{r} G^{\circ} = - 2.303 R T lo g_{10} K$

$Substitution$

$Isolating lo g_{10} K$

$Calculating K$

$Initial State$

$Equilibrium State$

$Equilibrium Expression$

$Solving for x$

$Final Algebra$

$Amount of B$

$What if?$

The Sigma Insight: Law of Mass Action

The Thermodynamic Bridge

Decoding the Math

The ICE Table

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; } A(g)⇌B(g)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δr​G∘=−2.303RTlog10​K

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolating log10​K

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

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

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

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

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

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

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

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