Q.PCl5, PCl3 and Cl2 are at equilibrium at 500 K and having concentration 1.59 M PCl3, 1.59 M Cl2 and 1.41 M PCl5. Calculate Kc for the reaction, PCl5 ⇌ PCl3 + Cl2.
Concept understanding — Equilibrium Constant Calculation
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.
- Ignoring pure solids and liquids: Their concentrations are constant and are absorbed into K — they do not appear in the expression. For example, in CaCO3(s)⇌CaO(s)+CO2(g), Kc=[CO2].
- Confusing Kc and Kp: Kp uses partial pressures (in atm or bar) instead of concentrations. The form is identical, but the numerical value differs unless Δn=0.
The Big Picture
K is a thermodynamic fingerprint of a reaction at a given temperature. It tells you:
- Direction: Compare Q (the reaction quotient, same formula but with any concentrations) to K. If Q<K, the reaction moves forward. If Q>K, it moves backward.
- Extent: Large K → products favoured; small K → reactants favoured.
- Temperature dependence: Use Le Chatelier's principle or the van't Hoff equation (for advanced problems).
Once you see K as a ratio of "what's made" to "what's left" at equilibrium, the calculations become straightforward — just careful algebra with the right numbers.
Many students find this page while searching "Equilibrium Constant Calculation formula chemistry" or "Equilibrium Constant Calculation important questions and answers"; the concept sits firmly within the Class 11 Chemistry NCERT/CBSE syllabus. It's also a frequent building block for numericals in JEE Main, NEET and state CET Chemistry papers, so treating it as a one-time memorisation task rather than an understood idea tends to backfire later.
The key idea is that for a homogeneous gaseous equilibrium, Kc is the ratio of product concentrations to reactant concentrations, each raised to its stoichiometric coefficient.
Step 1 – Write the equilibrium expression
For PCl5⇌PCl3+Cl2,
Kc=[PCl5][PCl3][Cl2]
Step 2 – Substitute the given equilibrium concentrations
[PCl3]=1.59 M, [Cl2]=1.59 M, [PCl5]=1.41 M
Kc=1.41(1.59)(1.59)
Step 3 – Calculate
1.59×1.59=2.5281
2.5281÷1.41≈1.793
The value is 1.79 M (rounded to three significant figures).
For the decomposition PCl5⇌PCl3+Cl2, the equilibrium constant Kc is simply the product of product concentrations divided by the reactant concentration at equilibrium. Substituting the given values gives Kc=1.79.
The equilibrium constant tells us the ratio of products to reactants at a given temperature, once the reaction has settled into a dynamic balance. For a reaction like PCl5⇌PCl3+Cl2, the expression is straightforward: Kc=[PCl5][PCl3][Cl2]. Notice that the coefficients in the balanced equation are all 1, so each concentration appears with an exponent of 1 — no squares or cubes to worry about here.
The problem gives us the equilibrium concentrations directly. That means we don’t need an ICE table or any initial data; we just plug the numbers into the expression and compute.
- Write the equilibrium expression For the reaction PCl5⇌PCl3+Cl2,
Kc=[PCl5][PCl3][Cl2]
- Substitute the given equilibrium concentrations [PCl3]=1.59 M, [Cl2]=1.59 M, [PCl5]=1.41 M
Kc=1.41(1.59)(1.59)
-
Calculate the numerator
1.59×1.59=2.5281
-
Divide by the denominator
Kc=1.412.5281≈1.793
Rounding to three significant figures (since the given concentrations have three), we get Kc=1.79.
A common mistake is to invert the fraction — putting [PCl5] in the numerator. Remember: products over reactants, each raised to the power of its coefficient. Here, the coefficient is 1 for all species, so it’s just product concentrations divided by reactant concentration.
Notice that [PCl3] and [Cl2] are equal here. That’s not a coincidence — if the reaction starts from pure PCl5, the stoichiometry forces them to be equal at equilibrium. But even if they weren’t, the formula remains the same.
The value of Kc is 1.79.
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.
2x=0.316(0.04−x)=0.01265−0.316x⇒2.316x=0.01265⇒x=0.00546.
Equilibrium [NO]=2x=0.0109≈0.011 mol L−1.
✓Final answerThe correct option is (E).
- 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:
64=x2(0.08)2=x20.0064
x2=640.0064=1×10−4 ⇒ x=0.01M
So [H2]=[I2]=0.01M.
✓Final answerThe correct option is (B).
- 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.
- Changing total pressure (at constant T) also only shifts position (and here Δng=0 anyway).
- Only a change in temperature alters the value of K.
✓Final answerThe correct option is (E).
- 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
[XO]2=0.5×(4×10−3)(8×10−3)=0.5×32×10−6=16×10−6.
[XO]=16×10−6=4×10−3M.
✓Final answerThe correct option is (A).
- 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.
Kc=(RT)ΔnKp=(0.0831×1000)10.1662=83.10.1662=2×10−3 mol lit−1
✓Final answerThe correct option is (B).
- 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.
For N2+O2⇌2NO, Δng=2−2=0, so Kp=Kc(RT)0=Kc=0.5.
✓Final answerThe correct option is (A).
- 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.
- Multiplying coefficients by 2: K→(1/Kc)2.
Kc′=(41)2=161.
✓Final answerThe correct option is (E).
- 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.
For N2+3H2⇌2NH3, moles of gaseous product = 2, moles of gaseous reactant = 4, so Δng=2−4=−2. Substituting: KP=KC(RT)−2.
✓Final answerThe correct option is (B).
- 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:
Kc=(1.5×10−3)(2×10−3)(3×10−3)2=3×10−69×10−6=3.
✓Final answerThe correct option is (E).
- 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.
Comparing the given values: 4.8×10−3, 4.0×1031, 4.1×10−48, 57.0, and 4.64×10−3. The overwhelmingly largest is 4.0×1031, so the formation of HCl proceeds nearly to completion.
✓Final answerThe correct option is (B).
- 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:
Kc=(5×10−3)(0.10)3(1.5×10−2)2=(5×10−3)(1×10−3)2.25×10−4
=5×10−62.25×10−4=45.0 dm6mol−3
✓Final answerThe correct option is (C).
- 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
- (C) PCl5(g)⇌PCl3(g)+Cl2(g): Δng=2−1=+1
- (D) H2(g)+I2(g)⇌2HI(g): Δng=2−2=0
- (E) N2O4(g)⇌2NO2(g): Δng=2−1=+1
Only (D) gives Δng=0.
✓Final answerThe correct option is (D).
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