The relation between Kc and Kp for the reaction A (g) + B (g) ⇌ 2C (g) + D (g) is a. Kc = Ykp b. Kp = Kc2 c. K

Choose the correct option · Q10

Q. The relation between Kc and Kp for the reaction A

(g) + B

(g) ⇌ 2C

(g) + D

(g) is
a. Kc = Ykp
b. Kp = Kc2
c. Kc = 1/Kp
d. Kp/Kc = 1

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Concept understanding — Relationship between Kp and Kc

$K_P$ and $K_C$ describe the same equilibrium, so a fixed relationship connects them. Starting from the ideal gas equation $P=(n/V)RT$ and noting that active mass equals molar concentration $n/V$ , each species' partial pressure can be written as (concentration) $\times RT$ ; substituting this into the $K_P$ expression and comparing with $K_C$ gives

K_P=K_C(RT)^{\Delta n_g}

where $\Delta n_g$ is the difference between the total moles of gaseous products and the total moles of gaseous reactants in the balanced equation. Three regimes follow: $\Delta n_g=0$ gives $K_P=K_C$ exactly (e.g. $H_2+I_2\rightleftharpoons 2HI$ , or ammonia synthesis's own $K_C/K_P=(RT)^2$ inverse case where $\Delta n_g=-2$ gives $K_P=K_C(RT)^{-2}$ , so $K_C/K_P=(RT)^2$ ); $\Delta n_g>0$ gives $K_P>K_C$ (e.g. $PCl_5\rightleftharpoons PCl_3+Cl_2$ , where $\Delta n_g=+1$ , so $K_C/K_P=1/(RT)$ ); and $\Delta n_g<0$ gives $K_P<K_C$ (e.g. $2SO_2+O_2\rightleftharpoons 2SO_3$ ). …

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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 ANNUAL1 markMCQ
$Q. Kc/Kp for the reaction, N2(g) + 3H2(g) <=> 2NH3(g) is :$
(a) RT
(b) 1/RT
(c) (RT)^2
(d) sqrt(RT)
›Reveal solutionSolution
For N2(g) + 3H2(g) <=> 2NH3(g), the relation Kp = Kc(RT)^Δn with Δn = -2 gives Kc/Kp = (RT)^2.

The general relation between Kp (equilibrium constant in terms of partial pressures) and Kc (equilibrium constant in terms of molar concentrations) for a gas-phase reaction is:

Kp = Kc (RT)^Δn

where Δn = (total moles of gaseous products) - (total moles of gaseous reactants).

For the reaction N2(g) + 3H2(g) <=> 2NH3(g):

moles of gaseous products = 2 (from 2NH3) …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. The delta ng value for the reaction 2SO2(g) + O2(g) <=> 2SO3(g)$
(a) 2
(b) 1
(c) -2
(d) -1
›Reveal solutionSolution
Delta-ng for 2SO2(g) + O2(g) <=> 2SO3(g) equals (gaseous product moles) - (gaseous reactant moles) = 2 - 3 = -1.

Delta-ng is used to relate the equilibrium constants Kp and Kc via Kp = Kc(RT)^(Delta-ng), and is defined purely from the stoichiometric coefficients of GASEOUS species in the balanced chemical equation:

Delta-ng = (sum of stoichiometric coefficients of gaseous products) - (sum of stoichiometric coefficients of gaseous reactants)

For the reaction 2SO2(g) + O2(g) <=> 2SO3(g):

Gaseous reactants: 2 mol SO2 + 1 mol O2 = 3 mol total

Gaseous products: 2 mol SO3 = 2 mol total

Delta-ng = 2 - 3 = -1

…
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CBSE 2019Set ANNUAL1 markMCQ
$Q. When Δng is negative in chemical equilibrium reaction then:$
(a) Kp < Kc
(b) Kp = 1/Kc
(c) Kp = Kc(RT)^-ve
(d) Kp > Kc
›Reveal solutionSolution

For a negative Δng, Kp is smaller than Kc.

For a gaseous equilibrium, Kp and Kc are related by:

Kp = Kc(RT)^Δng

where Δng = (moles of gaseous products) − (moles of gaseous reactants).

When Δng is negative (fewer moles of gaseous product than reactant, e.g. N2 + 3H2 ⇌ 2NH3, Δng = 2−4 = −2), the exponent on (RT) is negative, so (RT)^Δng = 1/(RT)^|Δng|, which is a fraction less than 1 (since RT > 1 in the units normally used). Multiplying Kc by a factor less than 1 gives a smaller value, so:

Kp = Kc × (a fraction < 1) ⟹ Kp < Kc

…

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Q.The relation between Kc and Kp for the reaction A

Concept understanding — Relationship between Kp and Kc

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