Derive Kp and Kc value for the equilibrium reaction H2(g) + I2(g) <=> 2HI(g).

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Q. Derive Kp and Kc value for the equilibrium reaction H2(g) + I2(g) <=> 2HI(g).

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2024Subjective· 3mImportance★★★★★

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By the law of mass action, Kc = [HI]^2/([H2][I2]) and Kp = p(HI)^2/(p(H2)p(I2)); because this reaction has Delta-ng = 0, Kp equals Kc here.

For the reversible gas-phase reaction: H2(g) + I2(g) <=> 2HI(g)

Deriving Kc: By the law of mass action, the equilibrium constant in terms of molar concentrations is written as the ratio of the product of the equilibrium concentrations of the products (each raised to its stoichiometric coefficient) to the product of the equilibrium concentrations of the reactants (similarly raised):

Kc = [HI]^2 / ([H2][I2])

Deriving Kp: Since all species are gaseous, the equilibrium constant can equally be expressed in terms of their equilibrium partial pressures (each gas's partial pressure being proportional to its concentration via pV = nRT, i.e. p = (n/V)RT = [concentration] x RT):

Kp = (p(HI))^2 / (p(H2) x p(I2))

Relation between Kp and Kc: In general, Kp = Kc (RT)^(Delta-ng), where Delta-ng = (moles of gaseous products) - (moles of gaseous reactants).

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

CBSE 2025Set ANNUAL2 marks
$Q. The equilibrium concentrations of NH3, N2 and H2 are 1.8 x 10^-2 M, 1.2 x 10^-2 M and 3 x 10^-2 M respectively. Calculate the equilibrium constant for the formation of NH3 from N2 and H2.$
›Reveal solutionSolution

Using Kc = [NH3]^2 / ([N2][H2]^3) with the given equilibrium concentrations, Kc works out to 1.0 x 10^3 (mol^-2 L^2).

The reaction for formation of ammonia is:

N2(g) + 3H2(g) <=> 2NH3(g)

The equilibrium constant expression (law of mass action) is:

Kc = [NH3]^2 / ([N2] . [H2]^3)

Given:

[NH3] = 1.8 x 10^-2 M

[N2] = 1.2 x 10^-2 M

[H2] = 3 x 10^-2 M

Step 1 — numerator:

[NH3]^2 = (1.8 x 10^-2)^2 = 3.24 x 10^-4

Step 2 — denominator:

[H2]^3 = (3 x 10^-2)^3 = 27 x 10^-6 = 2.7 x 10^-5

[N2] x [H2]^3 = (1.2 x 10^-2) x (2.7 x 10^-5) = 3.24 x 10^-7

…

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CBSE 2022Set ANNUAL2 marks
$Q. Give a balanced chemical equation for the equilibrium reaction for which the equilibrium constant is given by expression Kc = [NH3]^4 [O2]^5 / ([NO]^4 [H2O]^6).$
›Reveal solutionSolution
In an equilibrium constant expression, the exponents are the stoichiometric coefficients of that species, with products in the numerator and reactants in the denominator; reading Kc = [NH3]^4[O2]^5 / ([NO]^4[H2O]^6) this way and checking atom balance gives 4NO + 6H2O ⇌ 4NH3 + 5O2.

For a general equilibrium a A + b B ⇌ c C + d D, the equilibrium constant is Kc = [C]^c[D]^d / ([A]^a[B]^b) — products (right side of the equation) appear in the numerator, reactants (left side) in the denominator, each raised to its stoichiometric coefficient.

Here, Kc = [NH3]^4 [O2]^5 / ([NO]^4 [H2O]^6), so:

Products: 4 NH3 and 5 O2

Reactants: 4 NO and 6 H2O

Proposed equation: 4NO + 6H2O ⇌ 4NH3 + 5O2

Check atom balance: …
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Q.Derive Kp and Kc value for the equilibrium reaction H2(g) + I2(g) <=> 2HI(g).

Concept understanding — Equilibrium Constant Kc and Kp

Q.Derive Kp and Kc value for the equilibrium reaction H2(g) + I2(g) <=> 2HI(g).

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