If the equilibrium constant for A B + C is K_eq^(1) and that of B + C P is K_eq^(2), the equ

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

Because equilibrium constants are products of concentrations. When you add reactions, the products of the first become reactants of the second, and multiplying their $K$ values naturally cancels out these intermediate species, leaving the overall $K$.

Q: What happens to $K$ if we multiply a reaction by a number?

If you multiply a reaction by a stoichiometric coefficient $n$, the new equilibrium constant becomes $K^n$.

Q: Can we just add the $K$ values like we add enthalpy ($\Delta H$)?

No! Enthalpy is an additive property (state function), but $K$ is related to $\Delta G^\circ$ exponentially ($\Delta G^\circ = -RT \ln K$). Adding $\Delta G^\circ$ values results in multiplying the $K$ values due to the properties of logarithms.

Q: What if the intermediate species don't fully cancel out?

Then the overall reaction will still contain those species, and the multiplied $K$ value will correctly reflect their remaining stoichiometric coefficients in the final expression.

Q: Does this rule apply to $K_p$ as well as $K_c$?

Yes, the exact same rules apply to $K_p$, $K_a$, $K_b$, $K_{sp}$, and any other equilibrium constant.

Question

If the equilibrium constant for $A \rightleftharpoons B + C$ is $K_{eq}^{(1)}$ and that of $B + C \rightleftharpoons P$ is $K_{eq}^{(2)}$, the equilibrium constant for $A \rightleftharpoons P$ is

Options:

(A) $K_{eq}^{(1)} / K_{eq}^{(2)}$

(B) $K_{eq}^{(1)} + K_{eq}^{(2)}$

(C) $K_{eq}^{(1)} K_{eq}^{(2)}$

(D) $K_{eq}^{(2)} - K_{eq}^{(1)}$

Sigma Percentile · Accepted Answer

The beauty of chemical equilibrium lies in its mathematical elegance. When multiple reactions occur in sequence, their equilibrium constants don't just sit in isolation—they interact in a very predictable and logical way. Let's dive into this problem and see how intermediate steps combine to give us the overall picture.

### Analyzing the Setup

We are given two sequential reactions. In the first step, reactant $A$ decomposes or reacts to form an intermediate mixture of $B$ and $C$.

The equilibrium constant for this step is defined as:
$$K_{eq}^{(1)} = \frac{[B][C]}{[A]}$$

In the second step, these intermediates $B$ and $C$ immediately react further to form the final product $P$.

The equilibrium constant for this subsequent step is:
$$K_{eq}^{(2)} = \frac{[P]}{[B][C]}$$

Our ultimate goal is to find the equilibrium constant, let's call it $K_{eq}$, for the direct, overall transformation from $A$ to $P$.

The expression for this target reaction is:
$$K_{eq} = \frac{[P]}{[A]}$$

### The Master Equation

In thermodynamics and chemical equilibrium, there is a fundamental principle: **when you add two chemical equations together to get a net equation, you must multiply their equilibrium constants.**

Let's see why this happens. If we add our two given reactions:
$$(A ightleftharpoons B + C) + (B + C ightleftharpoons P)$$

The species $B$ and $C$ appear on both the reactant and product sides of the combined equation. Just like in an algebraic equation, they cancel out perfectly, leaving us with our target reaction:
$$A ightleftharpoons P$$

### Final Calculation

Now, let's apply the mathematical rule to the equilibrium constants. We multiply $K_{eq}^{(1)}$ and $K_{eq}^{(2)}$:
$$K_{eq}^{(1)} 	imes K_{eq}^{(2)} = \left( \frac{[B][C]}{[A]} ight) 	imes \left( \frac{[P]}{[B][C]} ight)$$

Notice how the concentration terms $[B]$ and $[C]$ are in the numerator of the first term and the denominator of the second term. They cancel out beautifully!
$$K_{eq}^{(1)} 	imes K_{eq}^{(2)} = \frac{[P]}{[A]}$$

And what is $\frac{[P]}{[A]}$? It is exactly the expression for our target equilibrium constant, $K_{eq}$.

Therefore, we have rigorously proven that:
$$K_{eq} = K_{eq}^{(1)} 	imes K_{eq}^{(2)}$$

This simple yet powerful rule saves us from doing complex derivations every time. Remember: **Add the reactions, multiply the K's!**

Animated Solution for Chemistry - Chemical Equilibrium: If the equilibrium constant for $A ⇌ B + C$ is $K_{e q}^{(1)}$ and that of $B + C ⇌ P$ is $K_{e q}^{(2)}$ , the equilibrium constant for $A ⇌ P$ is

Select Answer:

Visualized Solution

$A ⇌ B + C$

$B + C ⇌ P$

$A ⇌ P$

$Adding Reactions$

$K_{e q}^{(1)} \times K_{e q}^{(2)}$

$K_{e q} = K_{e q}^{(1)} \times K_{e q}^{(2)}$

$Key Takeaways$

The Sigma Insight: Law of Mass Action

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; } A⇌B+C

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Keq(1)​×Keq(2)​

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

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