What is K c for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60

Exercises · 6.2

Q. What is K c for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60M, [O2] = 0.82M and [SO3] = 1.90M ? 2SO2(g) + O2(g) ⇌ 2SO3(g)

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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

$A + B \rightleftharpoons C + D$ 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 $10^6$ ), 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 \rightleftharpoons cC + dD

the equilibrium constant $K_c$ (for concentrations in mol/L) is:

K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}

where $[X]$ means the equilibrium concentration of species X in moles per litre.

Important

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 $K_c$ ; $K_p$ 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:

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

At 500°C, equilibrium concentrations are:

$[N_2] = 0.50\ \text{M}$ , $[H_2] = 1.50\ \text{M}$ , $[NH_3] = 0.20\ \text{M}$

Calculate $K_c$ .

Solution:

Write the expression:

K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}

Substitute:

K_c = \frac{(0.20)^2}{(0.50)(1.50)^3} = \frac{0.04}{0.50 \times 3.375} = \frac{0.04}{1.6875} \approx 0.0237

Note

The units cancel because the numerator and denominator both have units of (mol/L) $^2$ and (mol/L) $^4$ respectively — but by convention, $K_c$ 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 $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ , $K_c = [CO_2]$ .
Confusing $K_c$ and $K_p$ : $K_p$ uses partial pressures (in atm or bar) instead of concentrations. The form is identical, but the numerical value differs unless $\Delta n = 0$ .

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.

Write the equilibrium expression from the balanced equation, substitute the given concentrations with proper stoichiometric exponents, and calculate. $K_c = 12.2$

The equilibrium constant $K_c$ captures the ratio of product to reactant concentrations at equilibrium, with each species raised to the power of its stoichiometric coefficient. It tells us where the equilibrium "sits"—whether products or reactants are favored. For any reaction, once you know the balanced equation, the form of $K_c$ is fixed by the stoichiometry.

For the reaction:

2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)

the equilibrium constant expression is:

K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}

Notice that products go in the numerator, reactants in the denominator, and each concentration is raised to the power matching its coefficient in the balanced equation.

Step-by-step calculation:

Identify the equilibrium concentrations from the problem:
- $[\text{SO}_2] = 0.60 \, \text{M}$
- $[\text{O}_2] = 0.82 \, \text{M}$
- $[\text{SO}_3] = 1.90 \, \text{M}$
Write the $K_c$ expression using the stoichiometric coefficients:

K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}

Substitute the equilibrium concentrations into the expression:

K_c = \frac{(1.90)^2}{(0.60)^2 \times (0.82)}

Calculate the numerator:

(1.90)^2 = 3.61

Calculate the denominator:

(0.60)^2 = 0.36

0.36 \times 0.82 = 0.2952

Divide to find $K_c$ :

K_c = \frac{3.61}{0.2952} \approx 12.23

Rounding to three significant figures (matching the precision of the given data):

K_c \approx 12.2

Tip

Always check that your $K_c$ expression matches the balanced equation exactly—coefficients become exponents, and only aqueous and gaseous species appear (pure solids and liquids are omitted).

✓Final answer

\boxed{K_c = 12.2}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 24 on this concept.

CBSE 2026Set sz1 markMCQ
$Q. Select the correct one: The rate at which a substance reacts, depends on its:$
(a) Active mass
(b) Molecular mass
(c) Equivalent mass
(d) Total volume
›Reveal solutionSolution
The Law of Mass Action states that the rate of a chemical reaction is proportional to the product of the active masses (molar concentrations) of the reactants, so the rate depends on active mass.

The Law of Mass Action (proposed by Guldberg and Waage) states that the rate at which a substance reacts is directly proportional to its active mass, where active mass is defined as the molar concentration of the substance (expressed in mol/L, i.e. [reactant]).

For a general reaction aA + bB -> products, the rate is proportional to [A]^a[B]^b — the exponents being the active masses raised to powers related to the stoichiometry (for elementary reactions).

Molecular mass and equivalent mass are fixed properties of a substance that do not change with how much of it is dissolved, so they cannot govern how fast a reaction proceeds.

Total volume alone (without knowing how much substance is dissolved in it) also says nothing about reaction rate; what matters is concentration (amount per unit volume), i.e. active mass.

✓Final answer
$The correct option is (a) Active mass.$
CBSE 2026Set ANNUAL1 mark
$Q. Give the name of mixture of reactants and products in the equilibrium state.$
›Reveal solutionSolution

At chemical equilibrium, the combined mixture of reactants and products present (in constant, unchanging proportions) is called the equilibrium mixture.

In a reversible reaction, as the reaction proceeds, the forward reaction rate decreases and the reverse reaction rate increases, until both become equal — this is the state of dynamic equilibrium. At this point, the concentrations of all reactants and products stop changing (though the forward and reverse reactions continue to occur at equal rates). The mixture of unreacted reactants together with the products formed, present at this stage, is termed the equilibrium mixture.

✓Final answer
$It is called the equilibrium mixture.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Who enunciated the law of mass action ?$
(a) Gulberg and Waage
(b) Bodenstein
(c) Berthelot
(d) Graham
›Reveal solutionSolution

The law of mass action was given by Guldberg and Waage.

Cato Guldberg and Peter Waage (1864) stated the law of mass action: the rate of a chemical reaction is proportional to the product of the active masses (molar concentrations) of the reactants, each raised to the power of its coefficient in the balanced equation. This law is the basis of the equilibrium constant expression.

✓Final answer
$(A) Guldberg and Waage.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. For which of the following reactions Kp > Kc ?$
(a) N2(g) + 3H2(g) ⇌ 2NH3(g)
(b) H2(g) + I2(g) ⇌ 2HI(g)
(c) PCl3(g) + Cl2(g) ⇌ PCl5(g)
(d) 2SO3(g) ⇌ 2SO2(g) + O2(g)
›Reveal solutionSolution
Kp > Kc requires Δn(gas) > 0; only 2SO3 ⇌ 2SO2 + O2 satisfies this.

The relation is Kp = Kc(RT)^Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants). Kp > Kc needs Δn > 0.

(A) N2 + 3H2 ⇌ 2NH3: Δn = 2 − 4 = −2 → Kp < Kc.

(B) H2 + I2 ⇌ 2HI: Δn = 2 − 2 = 0 → Kp = Kc.

(C) PCl3 + Cl2 ⇌ PCl5: Δn = 1 − 2 = −1 → Kp < Kc.

(D) 2SO3 ⇌ 2SO2 + O2: Δn = 3 − 2 = +1 → Kp > Kc.

✓Final answer
$(D) 2SO3(g) ⇌ 2SO2(g) + O2(g).$
CBSE 2026Set ANN1 markMCQ
$Q. Equilibrium constant (Kc) of a reaction depends on :$
(a) temperature
(b) pressure
(c) catalyst
(d) initial concentration
›Reveal solutionSolution
Kc depends only on temperature; it is unaffected by pressure, catalyst or initial concentration. Answer: (a).

The equilibrium constant is a fixed number for a given reaction at a given temperature.

Changing pressure or initial concentration shifts the position of equilibrium but does NOT change Kc.

A catalyst only speeds up attainment of equilibrium; it does not change Kc.

Only a change in temperature alters Kc (because it changes the rate constants of forward and reverse reactions differently).

✓Final answer
$(a) Temperature.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. In which of the following is Kp less than Kc?$
(a) PCl5 ⇌ PCl3 + Cl2
(b) H2 + Cl2 ⇌ 2HCl
(c) 2SO2 + O2 ⇌ 2SO3
(d) All of these
›Reveal solutionSolution
Kp < Kc only for the reaction where the number of gas moles decreases (Δn negative): 2SO2 + O2 ⇌ 2SO3.

The relation is Kp = Kc(RT)^Δn, where Δn = (moles of gaseous products) - (moles of gaseous reactants). Since RT > 1 (in atm-litre units, for reasonable temperatures), Kp > Kc when Δn is positive, Kp = Kc when Δn = 0, and Kp < Kc when Δn is negative.

Checking each reaction:

PCl5 ⇌ PCl3 + Cl2: Δn = 2 - 1 = +1, so Kp > Kc.

H2 + Cl2 ⇌ 2HCl: Δn = 2 - 2 = 0, so Kp = Kc.

2SO2 + O2 ⇌ 2SO3: Δn = 2 - 3 = -1, so Kp < Kc.

Only the third reaction has Kp < Kc.

✓Final answer
$(C) 2SO2 + O2 ⇌ 2SO3.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. If 64 gram of Hydrogen iodide are dissolved in 2 litre of water then its active mass will be$
(a) 0.5
(b) 0.25
(c) 1.0
(d) 2.5
›Reveal solutionSolution

64 g of HI in 2 L of water gives an active mass (concentration) of 0.25 mol/L.

Molar mass of HI = 1 + 127 = 128 g/mol.

Moles of HI = 64 g / 128 g mol^-1 = 0.5 mol.

Active mass (molar concentration) = moles / volume(L) = 0.5 mol / 2 L = 0.25 mol/L.

✓Final answer
$(B) 0.25.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. For the following chemical reaction PCl3(g) + Cl2(g) ⇌ PCl5(g). The value of Kc at 250°C is 26 then the value of Kp at this temperature will be$
(a) 0.61
(b) 0.57
(c) 0.83
(d) 0.46
›Reveal solutionSolution

For PCl3(g) + Cl2(g) ⇌ PCl5(g) at 250°C, Kp ≈ 0.61.

Δn = (moles of gaseous product) - (moles of gaseous reactants) = 1 - 2 = -1.

T = 250°C + 273 = 523 K. R = 0.0821 L atm K^-1 mol^-1.

RT = 0.0821 x 523 = 42.94.

Kp = Kc(RT)^Δn = 26 x (42.94)^-1 = 26 / 42.94 = 0.605 ≈ 0.61.

✓Final answer
$(A) 0.61.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. For a reversible reaction A + B (equilibrium arrows) C + D, the equilibrium constant Kc is equal to$
(a) [A][B] / [C][D]
(b) [C][A] / [B][D]
(c) [C][D] / [B][A]
(d) [A][D] / [B][C]
›Reveal solutionSolution

Kc = (concentrations of products, each raised to its coefficient) / (concentrations of reactants, each raised to its coefficient) — for A + B <=> C + D, that's [C][D]/[A][B].

For a general reversible reaction at equilibrium:

aA + bB <=> cC + dD

The law of mass action gives the equilibrium constant expression as:

Kc = [C]^c [D]^d / ([A]^a [B]^b)

Here the reaction is A + B <=> C + D (all coefficients = 1), so:

Kc = [C][D] / [A][B]

Comparing to the given options: option (c), [C][D]/[B][A], is algebraically identical to [C][D]/[A][B] — multiplying [B] and [A] together in either order gives the same denominator. The other options place a product concentration incorrectly in the numerator alongside a reactant (b) or invert the whole ratio, i.e. put reactants over products (a), or mix a reactant with a product in the numerator (d) — all of which violate the products-over-reactants rule.

✓Final answer
$(c) [C][D] / [B][A] = Kc.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. If the value of Kc for the equilibrium 2NOCl(g) <=> 2 NO(g) + Cl2(g) is 9.0 x 10^-4 mol L^-1 then numerical value of Kc for the equilibrium NOCl(g) <=> NO(g) + 1/2 Cl2(g) will be.................$
(a) 4.5 x 10^-4
(b) 3.0 x 10^-4
(c) 4.5 x 10^-2
(d) 3.0 x 10^-2
›Reveal solutionSolution

Halving all coefficients of a balanced equation converts K to sqrt(K). Since the given Kc = 9.0x10^-4 is for the doubled reaction, the halved reaction's Kc is sqrt(9.0x10^-4) = 3.0x10^-2.

For the reaction 2NOCl(g) <=> 2NO(g) + Cl2(g):

Kc = [NO]^2[Cl2] / [NOCl]^2 = 9.0x10^-4

The second reaction, NOCl(g) <=> NO(g) + 1/2 Cl2(g), is exactly half of the first reaction (all coefficients divided by 2).

Rule: if a reaction is divided by a factor n, the new equilibrium constant K' = K^(1/n).

Here n = 2, so:

K' = (Kc)^(1/2) = (9.0x10^-4)^(1/2)

sqrt(9.0x10^-4) = sqrt(9.0) x sqrt(10^-4) = 3.0 x 10^-2

✓Final answer
$The correct option is (d) 3.0x10^-2.$
CBSE 2025Set ANNUAL1 mark
$Q. Case study: A chemist while studying a number of equilibria found that there was a relationship between Kp and Kc. He tested this relation upon various equilibria at different temperatures and the relation was found to be true. He further noticed that for certain equilibria Kp and Kc were equal, however it was not always true. Based on above study answer the following question:$
(b) Under what condition Kp and Kc are equal?
›Reveal solutionSolution

From Kp = Kc(RT)^delta-n, the two constants become equal exactly when the exponent delta-n is zero, i.e., when there is no change in the total number of moles of gas between reactants and products.

From the derived relation, Kp = Kc(RT)^delta-n, where delta-n = (moles of gaseous products) - (moles of gaseous reactants).

If delta-n = 0 (i.e., the number of moles of gaseous species is the same on both sides of the balanced equation), then (RT)^delta-n = (RT)^0 = 1, and the relation simplifies to:

Kp = Kc

Example: H2(g) + I2(g) <=> 2HI(g) has delta-n = 2 - (1+1) = 0, so Kp = Kc for this equilibrium.

✓Final answer
$Kp and Kc are equal exactly when delta-n = 0 — that is, when the total number of moles of gaseous products equals the total number of moles of gaseous reactants.$
CBSE 2025Set ANNUAL1 mark
$Q. Case study: A chemist while studying a number of equilibria found that there was a relationship between Kp and Kc. He tested this relation upon various equilibria at different temperatures and the relation was found to be true. He further noticed that for certain equilibria Kp and Kc were equal, however it was not always true. Based on above study answer the following question:$
(c) Write the relation between Kp and Kc for the equilibrium 2H2(g) + O2(g) <=> 2H2O(g)
›Reveal solutionSolution

With delta-n = 2 - (2+1) = -1 for this reaction, the general relation Kp = Kc(RT)^delta-n becomes Kp = Kc(RT)^-1, i.e., Kp = Kc/(RT).

For the equilibrium: 2H2(g) + O2(g) <=> 2H2O(g)

Count gaseous moles:

Reactants: 2 mol H2 + 1 mol O2 = 3 mol of gas

Products: 2 mol H2O = 2 mol of gas

delta-n = (moles of gaseous products) - (moles of gaseous reactants) = 2 - 3 = -1

Using the general relation Kp = Kc(RT)^delta-n:

Kp = Kc(RT)^(-1) = Kc / (RT)

So for this reaction, Kp is smaller than Kc (since dividing by RT, a positive quantity greater than 1 in typical units, reduces the numerical value) — this makes sense because the reaction reduces the total number of gas moles, so the pressure-based constant differs from the concentration-based constant by a factor of RT.

✓Final answer
$Kp = Kc/(RT), since delta-n = -1 for 2H2(g) + O2(g) <=> 2H2O(g).$

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Q.What is K c for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60M, [O2] = 0.82M and [SO3] = 1.90M ? 2SO2(g) + O2(g) ⇌ 2SO3(g)

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