In the figure shown below, reactant A (represented by square) is in equilibrium with product B (repr

Q: Why can we just use the number of particles instead of their actual concentrations in moles per liter?

Because both the reactants (squares) and products (circles) are in the exact same container, they share the same volume $V$. When you write the ratio $\frac{n_B/V}{n_A/V}$, the volume $V$ cancels out perfectly, leaving just the ratio of the number of particles!

Q: I counted 10 circles instead of 11. Does that make my answer wrong?

Not at all! If you use 10 circles, your calculation would be $K = \frac{10}{6} \approx 1.66$. The closest integer option provided in the question is still $2$. The slight ambiguity in the drawing doesn't change the final correct option.

Q: What if the reaction was $A \rightleftharpoons 2B$? How would the calculation change?

If the stoichiometry was different, the equilibrium expression would change. For $A \rightleftharpoons 2B$, the constant would be $K = \frac{[B]^2}{[A]}$. You would then need to square the concentration of the circles, which would require knowing the actual volume of the container.

Q: Does the position or arrangement of the squares and circles in the box matter?

No, the spatial arrangement is completely irrelevant for calculating the equilibrium constant. In a real gas or solution, particles are constantly moving randomly. Only the total count of each type of particle at equilibrium matters.

Q: Why is the equilibrium constant rounded to the nearest integer?

In multiple-choice questions, especially those based on visual approximations, the options are often simplified. Since $1.83$ is very close to $2$, and the other options ($1, 4, 8$) are far off, $2$ is the only logical choice.

Question

In the figure shown below, reactant A (represented by square) is in equilibrium with product B (represented by circle). The equilibrium constant is

Options:

(A) 4

(B) 2

(C) 8

(D) 1

Sigma Percentile · Accepted Answer

### Visualizing Chemical Equilibrium

Imagine you have a microscopic camera that lets you peek inside a reaction vessel. Instead of abstract chemical formulas, you see the actual molecules bouncing around. In this fascinating problem, we are given exactly that: a visual snapshot of a chemical system in a state of dynamic equilibrium.

The reactant, let's call it $A$, is represented by the **squares**. The product, $B$, is represented by the **circles**. The reaction is a simple transformation: 
$$A ightleftharpoons B$$

Our goal is to determine the equilibrium constant, $K$, just by looking at this picture.

### The Master Equation

How do we connect a drawing to a mathematical constant? The equilibrium constant $K$ is defined as the ratio of the concentration of the products to the concentration of the reactants at equilibrium.

Mathematically, this is written as:
$$K = \frac{[B]}{[A]}$$

Since all these particles are in the same container, they share the same volume. Therefore, the ratio of their concentrations is exactly equal to the ratio of their numbers.

$$K = \frac{	ext{Number of circles}}{	ext{Number of squares}}$$

### Counting the Particles

Now comes the fun part—counting! If we carefully scan the container, we can count the number of squares. There are exactly **$6$ squares**.

Next, we count the circles. Depending on how closely you look at the grid, you might count $10$ or $11$ circles. The official data for this specific problem takes the number of circles as **$11$**.

Let's substitute these values into our equilibrium expression:
$$K = \frac{11}{6}$$

### Final Calculation

All that's left is a simple division.

$$K \approx 1.833$$

Looking at our options, we need to find the closest integer value. The value $1.833$ is closest to **$2$**.

Therefore, the equilibrium constant for this reaction is approximately $2$. This visual approach beautifully bridges the gap between abstract mathematical formulas and the physical reality of molecules in a container.

Animated Solution for Chemistry - Chemical Equilibrium: In the figure shown below, reactant A (represented by square) is in equilibrium with product B (represented by circle). The equilibrium constant is

Select Answer:

Visualized Solution

$A ⇌ B$

$K = \frac{[ B ]}{[ A ]}$

$Counting Particles$

$K = \frac{11}{6}$

$K \approx 2$

$The Way Forward$

The Sigma Insight: Law of Mass Action

Visualizing Chemical Equilibrium

The Master Equation

Counting the Particles

Final Calculation

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Animated Solution for Chemistry - Chemical Equilibrium: In the figure shown below, reactant A (represented by square) is in equilibrium with product B (represented by circle). The equilibrium constant is

Select Answer:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K=[A][B]​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K=611​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K≈2

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