For the following three reactions I, II and III, equilibrium constants are given
I. CO(g) + H2O(g)

Q: Why do we multiply the equilibrium constants when adding reactions?

When you add reactions, you are essentially multiplying their equilibrium constant expressions (which are fractions of products over reactants). The intermediate species that cancel out in the chemical equation also cancel out in the numerator and denominator of the multiplied expressions, leaving the exact expression for the net reaction.

Q: What happens to the equilibrium constant if we reverse a reaction?

If you reverse a chemical reaction, the products become reactants and vice versa. Therefore, the new equilibrium constant is the reciprocal of the original one ($1/K$).

Q: What if we multiply a reaction by a number 'n'?

When you multiply a chemical equation by a stoichiometric coefficient 'n', the new equilibrium constant becomes the original constant raised to the power of 'n' ($K^n$).

Q: Why did CO cancel out when adding the reactions?

Carbon monoxide (CO) appears as a reactant in Reaction I and as a product in Reaction II. When the two equations are added, CO appears on both sides of the equilibrium arrow. Just like in an algebraic equation, identical terms on opposite sides cancel each other out.

Q: Can we subtract reactions instead of adding them?

Yes! Subtracting a reaction is mathematically identical to adding its reverse. In that case, you would divide the equilibrium constants instead of multiplying them (since adding a reversed reaction means multiplying by $1/K$).

Question

For the following three reactions I, II and III, equilibrium constants are given
I. $CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g); K_1$
II. $CH_4(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g); K_2$
III. $CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g); K_3$
Which of the following relations is correct?

Options:

(A) $K_1\sqrt{K_2} = K_3$

(B) $K_2K_3 = K_1$

(C) $K_3 = K_1K_2$

(D) $K_3K_2^3 = K_1^2$

Sigma Percentile · Accepted Answer

### The Magic of Chemical Equations

Imagine you are a chemical architect, building complex reactions from simpler ones. In this problem, we are given three distinct chemical reactions along with their equilibrium constants, $K_1$, $K_2$, and $K_3$. Our mission is to find the hidden mathematical thread that connects them.

### Analyzing the Setup

Let's lay out our building blocks:

**Reaction I:** 
$$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)$$ 
with equilibrium constant $K_1$.

**Reaction II:** 
$$CH_4(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g)$$ 
with equilibrium constant $K_2$.

**Reaction III:** 
$$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g)$$ 
with equilibrium constant $K_3$.

At first glance, they might look like three independent processes. But in chemistry, reactions often behave like algebraic equations. We can add them, subtract them, or multiply them by scalars to generate new reactions.

### The Master Equation

Let's see what happens if we simply add **Reaction I** and **Reaction II** together.

On the reactant side (left-hand side), we gather everything:
$$CO(g) + H_2O(g) + CH_4(g) + H_2O(g)$$

On the product side (right-hand side), we do the same:
$$CO_2(g) + H_2(g) + CO(g) + 3H_2(g)$$

Now, just like in algebra, if a species appears on both sides of the arrow, it cancels out. Here, $CO(g)$ is present on both sides. Let's remove it. Also, we can group the similar terms together. We have two $H_2O(g)$ molecules on the left and a total of four $H_2(g)$ molecules on the right.

The resulting net reaction is:
$$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g)$$

Look closely at this net reaction. It is exactly **Reaction III**!

### Final Calculation

Now, we must apply a fundamental law of chemical equilibrium: **When two or more chemical reactions are added to yield a net reaction, the equilibrium constant of the net reaction is the product of the equilibrium constants of the individual reactions.**

Why does this happen? Because the equilibrium constant is a ratio of products to reactants. When you add reactions, you are multiplying these ratios, and the intermediate species (like $CO$ in our case) cancel out perfectly in the numerator and denominator.

Since Reaction III is the sum of Reaction I and Reaction II, its equilibrium constant $K_3$ must be the product of $K_1$ and $K_2$.

Therefore, the correct relation is:
**$$K_3 = K_1 \cdot K_2$$**

This elegant property allows chemists to determine the equilibrium constants of complex reactions without having to measure them directly, simply by breaking them down into known, simpler steps.

Select Answer:

Visualized Solution

$Analyzing the Given Reactions$

$Adding Reaction I and II$

$Relating the Equilibrium Constants$

The Sigma Insight: Law of Mass Action

The Magic of Chemical Equations

Analyzing the Setup

The Master Equation

Final Calculation

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Adding Reaction I and II

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