Q.At equilibrium, the concentrations of N2 = 3.0 × 10⁻³ M, O2 = 4.2 × 10⁻³ M and NO = 2.8 × 10⁻³ M in a sealed vessel at 800 K. What will be Kc for the reaction N2
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Equilibrium Constant Calculation: From Intuition to Precision
Imagine you're at a party where people can move between two rooms. Some people prefer the kitchen (more snacks), others prefer the living room (better music). After a while, the number of people in each room stops changing — not because everyone froze, but because the rate of people leaving the kitchen equals the rate of people entering it. The system is in dynamic equilibrium.
Chemical reactions work the same way. A reversible reaction like
A+B⇌C+D doesn't stop when it reaches equilibrium. Instead, the forward reaction (making C and D) and the reverse reaction (making A and B) happen at the same rate. The concentrations of A, B, C, and D become constant — not equal, but constant.
The equilibrium constant K is a number that tells you where this balance lies. It answers the question: At equilibrium, which side of the reaction is favoured?
The Intuitive Idea
Think of a seesaw. If K is very large (say 106), the equilibrium sits heavily on the product side — almost all A and B have turned into C and D. If K is very small (say 10−6), the opposite is true: hardly any product forms. If K is around 1, both sides have comparable amounts.
So K is a ratio — a comparison of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficients.
The Precise Statement
For a general reversible reaction at a given temperature:
aA+bB⇌cC+dD
the equilibrium constant Kc (for concentrations in mol/L) is:
Kc=[A]a[B]b[C]c[D]d
where [X] means the equilibrium concentration of species X in moles per litre.
The exponents come directly from the balanced chemical equation. If the coefficient of A is 2, you square its concentration. This is not optional — it's built into the definition.
What K Actually Depends On
K is constant at a given temperature. Change the temperature, and K changes. But K does not depend on:
- Initial concentrations
- Presence of a catalyst (catalysts speed up both directions equally)
- Pressure or volume changes (for Kc; Kp for gases has its own rules)
This is a critical exam point: if a problem gives you initial concentrations and asks for K, you must first find equilibrium concentrations — you cannot plug in initial values.
A Worked Example
Problem:
N2(g)+3H2(g)⇌2NH3(g)
At 500°C, equilibrium concentrations are:
[N2]=0.50 M, [H2]=1.50 M, [NH3]=0.20 M
Calculate Kc.
Solution:
Write the expression:
Kc=[N2][H2]3[NH3]2
Substitute:
Kc=(0.50)(1.50)3(0.20)2=0.50×3.3750.04=1.68750.04≈0.0237
The units cancel because the numerator and denominator both have units of (mol/L)2 and (mol/L)4 respectively — but by convention, Kc is reported without units. The numerical value is what matters.
Common Pitfalls
- Forgetting the exponents: A coefficient of 2 means square the concentration, not double it.
- Using initial concentrations: You must use equilibrium concentrations only. …
Concept: Equilibrium Constant Calculation
The equilibrium constant Kc relates the concentrations of products and reactants at equilibrium through the balanced equation. For the reaction NX2(g)+OX2(g)2NO(g), we write:
Kc=[NX2][OX2][NO]2
The exponent on [NO] is 2 because the stoichiometric coefficient of NO in the balanced equation is 2.
Substitute the given equilibrium concentrations: …
The equilibrium constant Kc is the ratio of product concentrations to reactant concentrations, each raised to their stoichiometric coefficients. For this reaction, Kc=0.622.
Understanding the Equilibrium Constant
The equilibrium constant Kc tells us the position of equilibrium — how far a reaction proceeds before the forward and reverse rates balance. It's defined purely from the balanced equation and the concentrations at equilibrium.
For any reaction aA+bB⇌cC+dD, we write:
Kc=[A]a[B]b[C]c[D]d
Products go in the numerator, reactants in the denominator, each raised to the power of its coefficient. The key insight: Kc is dimensionless in the way we use it here (though technically it has units that depend on the reaction stoichiometry), and it depends only on temperature.
Step-by-Step Calculation
1. Write the balanced equation and identify the Kc expression
The reaction is:
NX2(g)+OX2(g)2NO(g)
Notice that NO has a coefficient of 2. The equilibrium constant becomes:
Kc=[NX2][OX2][NO]2
2. Identify the equilibrium concentrations
We're given all three concentrations at equilibrium at 800 K:
- [NX2]=3.0×10−3M
- [OX2]=4.2×10−3M
- [NO]=2.8×10−3M
3. Substitute into the Kc expression
Kc=(3.0×10−3)(4.2×10−3)(2.8×10−3)2
4. Calculate the numerator
(2.8×10−3)2=7.84×10−6
5. Calculate the denominator …
Showing the 12 most recent of 25 on this concept.
- KEAM 2026Set eng-2026-04174 marksMCQQ.For the gas phase homogenous equilibrium N2(g)+O2(g)⇌2NO(g), KC= 0.1 at 1500K. If the initial concentrations of N2(g) and O2(g) are each 0.04 mol L−1 what is the equilibrium concentration of NO(g)? (A) 0.100 mol L−1 (B) 0.0100mol L−1 (C) 0.022 mol L−1 (D) 0.02 mol L−1 (E) 0.011 mol L−1
›Reveal solutionSolution
Because Δn=0 the equilibrium expression is a perfect square, so taking Kc linearises it; [NO]≈0.011 mol/L.
For N2+O2⇌2NO, let x mol/L of each reactant react, forming 2x of NO:
Kc=[N2][O2][NO]2=(0.04−x)2(2x)2=0.1.
Take the square root:
0.04−x2x=0.1=0.316. …
- KEAM 2026Set eng-2026-04184 marksMCQQ.A certain amount of H2(g) and I2(g) are sealed in a 4L container and kept at 600K to attain equilibrium. KC for the reaction, H2(g)+I2(g)⇌2HI(g) at 600K is 64. If the equilibrium concentration of HI(g) is 0.08M, what are the equilibrium concentrations of H2(g) and I2(g) at the same temperature? (A) 0.02 M (B) 0.01M (C) 0.03 M (D) 0.04 M (E) 0.002 M
›Reveal solutionSolution
Setting [H2]=[I2]=x in KC=[HI]2/([H2][I2])=64 gives x=0.01M.
For H2+I2⇌2HI:
KC=[H2][I2][HI]2
Starting from equal amounts of H2 and I2, at equilibrium [H2]=[I2]=x. With [HI]=0.08M and KC=64: …
- KEAM 2026Set eng-2026-04184 marksMCQQ.Consider the following equilibrium A(g)+B(g)⇌C(g)+D(g). The equilibrium constant will be altered when (A) reactant 'A' is added to the system at constant temperature (B) product 'D' is added to the system at constant temperature (C) total pressure is changed at constant temperature (D) both 'A' and 'D' are added to the system at constant temperature (E) temperature is changed
›Reveal solutionSolution
The equilibrium constant is a function of temperature only; concentration or pressure changes shift the equilibrium but leave K unchanged.
The equilibrium constant K is fixed at a given temperature.
- Adding A, D, or both (at constant T) shifts the equilibrium position (Le Chatelier) but K stays the same. …
- KEAM 2026Set eng-2026-04194 marksMCQQ.For the equilibrium, X2(g) + O2(g) = 2XO(g), the equilibrium concentrations of X2(g) and O2(g) are 4 x 10−3 M and 8 x 10−3 M respectively. What is the equilibrium concentration of XO(g)? (Equilibrium constant Kc = 0.5) (A) 4×10−3 M (B) 6×10−3 M (C) 5×10−3 M (D) 2×10−3 M (E) 8×10−3 M
›Reveal solutionSolution
From Kc=[X2][O2][XO]2, solving gives [XO]=4×10−3M.
For X2+O2⇌2XO:
Kc=[X2][O2][XO]2=0.5.
So …
- KEAM 2026Set eng-2026-04204 marksMCQQ.For the equilibrium, 2A(g)⇌2B(g)+C(g), the value of the equilibrium constant, Kp is 0.1662 atm at 1000 K. The value of Kc for the equilibrium at the same temperature is (R = 0.0831 lit atm mol−1) (A) 1×10−3 mol lit−1 (B) 2×10−3 mol lit−1 (C) 0.4×10−3 mol lit−1 (D) 1.6×10−3 mol lit−1 (E) 0.6×10−3 mol lit−1
›Reveal solutionSolution
Kp=Kc(RT)Δn with Δn=+1.
For 2A⇌2B+C: Δng=(2+1)−2=1. …
- KEAM 2026Set eng-2026-04224 marksMCQQ.The reaction between N2 and O2 is carried out in a sealed vessel at 300 K. At equilibrium, [N2]=2×10−3 M, [O2]=4×10−3 M and [NO]=2×10−3 M. What is the value of Kp for the reaction N2(g)+O2(g)⇌2NO(g) at 300 K? (R = 0.08 bar lit mol−1 K) (A) 0.5 (B) 12 (C) 120 (D) 5 (E) 50
›Reveal solutionSolution
Kc=[N2][O2][NO]2=0.5, and since Δng=0, Kp=Kc(RT)Δn=0.5.
Kc=(2×10−3)(4×10−3)(2×10−3)2=8×10−64×10−6=0.5. …
- KEAM 2026Set pha-2026-0418F4 marksMCQQ.KC for the reaction A(g)+2B(g)⇌C(g)+2D(g) is 4.0 at 300 K. What is the value of K'c for the reaction, 2C(g)+4D(g)⇌2A(g)+4B(g) at 300 K? (A) 8.0 (B) 1/8 (C) 1/2 (D) 16 (E) 1/16
›Reveal solutionSolution
Reversing halves→inverts K and doubling squares it: Kc′=(1/4)2=1/16.
Reasoning
Given A+2B⇌C+2D, Kc=4.0.
The target 2C+4D⇌2A+4B is the original reaction reversed and multiplied by 2.
- Reversing: K→1/Kc=1/4. …
- KEAM 2026Set pha-2026-0419F4 marksMCQQ.For the homogenous equilibrium N2(g)+3H2(g)⇌2NH3(g), the equilibrium constants KP and KC are related as (A) KP=KC(RT)2 (B) KP=KC(RT)−2 (C) KP=KC(RT)−3 (D) KP=KC(RT)3 (E) KP=KC(RT)
›Reveal solutionSolution
KP=KC(RT)Δng with Δng=2−(1+3)=−2. …
- KEAM 2026Set pha-2026-0420F4 marksMCQQ.At equilibrium, [N2]=1.5×10−3M, [O2]=2×10−3M and [NO]=3×10−3M at 800 K in a closed vessel. The Kc for the equilibrium N2(g)+O2(g)⇌NO(g) at 800 K is (A) 1.0 (B) 0.3 (C) 2.0 (D) 4.0 (E) 3.0
›Reveal solutionSolution
Kc=[NO]2/([N2][O2])=9×10−6/3×10−6=3.
For the equilibrium N2(g)+O2(g)⇌2NO(g),
Kc=[N2][O2][NO]2.
Substituting the equilibrium concentrations: …
- KEAM 2025Set eng-2025-04254 marksMCQQ.Which of the following reaction proceeds nearly to completion? (Reaction — Kc value) (A) N2(g)+O2(g)⇌2NO(g) at 298 K — 4.8×10−3 (B) H2(g)+Cl2(g)⇌2HCl(g) at 300 K — 4.0×1031 (C) Decomposition of H2O into H2 and O2 at 500 K. — 4.1×10−48 (D) Reaction of H2 with I2 to give HI at 700 K. — 57.0 (E) Gas phase decomposition of N2O4 to NO2 at 298K. — 4.64×10−3
›Reveal solutionSolution
A reaction proceeds nearly to completion when its equilibrium constant is very large (Kc≫1). The largest Kc here is 4.0×1031 for H2+Cl2⇌2HCl.
The magnitude of Kc tells us the extent of a reaction:
- Kc≫1: products are strongly favoured (reaction nearly complete).
- Kc≪1: reactants are favoured (little reaction).
- Kc≈1: comparable amounts of reactants and products. …
- KEAM 2025Set eng-2025-04264 marksMCQQ.The following concentrations were obtained in the formation of NH3(g) from N2(g) and H2(g) at equilibrium at 500 K: [NH3]=1.5×10−2 M, [N2]=5×10−3 M and [H2]=0.10 M Calculate the equilibrium constant for the reaction (in dm6 mol−3) at 500 K. N2(g)+3H2(g)⇌2NH3(g) (A) 0.45 (B) 4.5 (C) 45.0 (D) 4.5×10−2 (E) 4.5×10−3
›Reveal solutionSolution
Substituting the equilibrium concentrations into Kc=[NH3]2/([N2][H2]3) gives 45.0.
For N2(g)+3H2(g)⇌2NH3(g),
Kc=[N2][H2]3[NH3]2
Substitute [NH3]=1.5×10−2, [N2]=5×10−3, [H2]=0.10: …
- KEAM 2025Set eng-2025-04274 marksMCQQ.In which of the following equilibrium KP=KC ? (A) CaCO3(s)⇌CaO(s)+CO2(g) (B) 2SO2(g)+O2(g)⇌2SO3(g) (C) PCl5(g)⇌PCl3(g)+Cl2(g) (D) H2(g)+I2(g)⇌2HI(g) (E) N2O4(g)⇌2NO2(g)
›Reveal solutionSolution
KP=KC(RT)Δng, so the two are equal only when the change in moles of gas Δng=0. That holds for the H2+I2⇌2HI equilibrium.
Reasoning
The relation between the two equilibrium constants is
KP=KC(RT)Δng
where Δng=(moles of gaseous products)−(moles of gaseous reactants). Setting KP=KC requires (RT)Δng=1, i.e. Δng=0.
Check each equilibrium (count only gaseous species):
- (A) CaCO3(s)⇌CaO(s)+CO2(g): Δng=1−0=+1
- (B) 2SO2(g)+O2(g)⇌2SO3(g): Δng=2−3=−1 …
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