What is K₃ for ½x₂+½y₂+½z₂⇌xyz given K₁ and K₂?

K₃=1.0×10⁻⁵. From ½×(reversed Reaction 1): K'=1/√(2.5×10⁵)=1/500=2×10⁻³. K₃=K'×K₂=2×10⁻³×5×10⁻³=10⁻⁵.

What happens to K when you halve a reaction?

K becomes K^(1/2) = √K. When reaction coefficients are multiplied by n, K becomes Kⁿ.

2xy⇌x₂+y₂ K₁=2.5×10⁵, xy+½z₂⇌xyz K₂=5×10⁻³ — find K₃ for ½x₂+½y₂+½z₂⇌xyz

JEERANKERS

ChemistryEquilibrium

QMCQChemical Equilibrium

Consider the following reactions in gaseous state:
$2\text{xy} \rightleftharpoons \text{x}_2 + \text{y}_2$,   $K_1 = 2.5\times10^5$
$\text{xy} + \tfrac{1}{2}\text{z}_2 \rightleftharpoons \text{xyz}$,   $K_2 = 5\times10^{-3}$

The value of $K_3$ for the equilibrium $\tfrac{1}{2}\text{x}_2 + \tfrac{1}{2}\text{y}_2 + \tfrac{1}{2}\text{z}_2 \rightleftharpoons \text{xyz}$ is:

A) $2.5\times10^{-3}$    B) $2.5\times10^{3}$    C) $1.0\times10^{-5}$    D) $5\times10^{-3}$

✅ Correct Answer

C) 1.0×10⁻⁵

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Solution

Manipulate Reaction 1 to get ½x₂ + ½y₂ → xy

Reverse Reaction 1: $\text{x}_2 + \text{y}_2 \rightleftharpoons 2\text{xy}$, $K = \dfrac{1}{K_1} = \dfrac{1}{2.5\times10^5}$

Multiply by ½ (take square root): $\tfrac{1}{2}\text{x}_2 + \tfrac{1}{2}\text{y}_2 \rightleftharpoons \text{xy}$, $K' = \sqrt{\dfrac{1}{K_1}} = \dfrac{1}{\sqrt{K_1}}$

$\sqrt{K_1} = \sqrt{2.5\times10^5} = \sqrt{250000} = 500$

$K' = \dfrac{1}{500} = 2\times10^{-3}$

Add to Reaction 2

Reaction 2 (unchanged): $\text{xy} + \tfrac{1}{2}\text{z}_2 \rightleftharpoons \text{xyz}$, $K_2 = 5\times10^{-3}$

Adding the two reactions gives target reaction:

$\tfrac{1}{2}\text{x}_2 + \tfrac{1}{2}\text{y}_2 \rightleftharpoons \text{xy}$ ($K'=2\times10^{-3}$)
$+\ \text{xy} + \tfrac{1}{2}\text{z}_2 \rightleftharpoons \text{xyz}$ ($K_2=5\times10^{-3}$)

$K_3 = K' \times K_2 = 2\times10^{-3} \times 5\times10^{-3} = 10\times10^{-6} = 1.0\times10^{-5}$

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Theory & Key Concepts

Manipulating Equilibrium Constants — Rules

There are three key rules for manipulating equilibrium constants corresponding to algebraic changes in chemical equations:

Rule 1 — Reverse the reaction: If a reaction is reversed, the new equilibrium constant = $1/K$ (reciprocal).

Rule 2 — Multiply reaction by n: If all stoichiometric coefficients are multiplied by a factor $n$, the new equilibrium constant = $K^n$. For half the reaction (n=½): $K^{1/2} = \sqrt{K}$.

Rule 3 — Add reactions: If two reactions are added, the overall equilibrium constant = product of the individual equilibrium constants: $K_{overall} = K_1 \times K_2$.

Physical meaning: These rules follow because $K = e^{-\Delta G°/RT}$ and $\Delta G°$ is additive (just like Hess's law for ΔH). Reversing a reaction negates $\Delta G°$ → reciprocal of $K$. Multiplying reaction by $n$ scales $\Delta G°$ by $n$ → $K^n$.

This problem step-by-step: $\sqrt{2.5\times10^5} = \sqrt{250000} = 500$. Reversed+halved Reaction 1 gives $K'=1/500=2\times10^{-3}$. Multiply with $K_2=5\times10^{-3}$: $K_3=10^{-5}$.

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Important Concepts

Reversing a Reaction — K becomes 1/K

Reaction 1: 2xy ⇌ x₂+y₂, K₁=2.5×10⁵. Reversed: x₂+y₂ ⇌ 2xy, K=1/K₁=1/(2.5×10⁵)=4×10⁻⁶. This is the Hess's law equivalent for equilibria.

Halving a Reaction — K becomes √K

When you multiply a reaction by ½ (halve all coefficients), the new K is K^(1/2)=√K. Here: ½×(x₂+y₂→2xy) gives K'=√(1/K₁)=1/√(2.5×10⁵)=1/500=2×10⁻³.

Adding Reactions — K values Multiply

When two reactions are added, K_total=K_1×K_2. Adding the modified Reaction 1 (K'=2×10⁻³) and Reaction 2 (K₂=5×10⁻³): K₃=2×10⁻³×5×10⁻³=10⁻⁵.

√(2.5×10⁵) = 500

2.5×10⁵=250000. √250000=√(2500×100)=50×10=500. Always simplify surds before computing — this avoids calculation errors in exam conditions.

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FAQs

What are the three K manipulation rules?

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(1) Reverse reaction → K becomes 1/K. (2) Multiply by n → K becomes Kⁿ. (3) Add reactions → K values multiply.

How does reversing Reaction 1 help?

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We need ½x₂+½y₂ on the left side. Reaction 1 has x₂+y₂ on the right. Reversing gives x₂+y₂ on left. Then halving gives ½x₂+½y₂ on left. Combined K'=1/√K₁.

What is √(2.5×10⁵)?

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√(250000)=500. So K'=1/500=2×10⁻³.

Final multiplication: K'×K₂?

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2×10⁻³ × 5×10⁻³ = 10×10⁻⁶ = 1.0×10⁻⁵.

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026 Chemistry Section A.

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