Derive mathematically value of kP for for A(g) + B(g) ⇌ C(g) + D(g)

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Q. Derive mathematically value of kP for for A(g) + B(g) ⇌ C(g) + D(g)

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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. $2SO_2+O_2\rightleftharpoons 2SO_3$ ).

This relation is the standard tool for converting a measured $K_P$ into $K_C$ or vice versa -- for example, $SrCO_3(s)\rightleftharpoons SrO(s)+CO_2(g)$ has $\Delta n_g=+1$ (only $CO_2$ is gaseous), so $K_C=K_P/(RT)$ ; at $K_P=2.2\times10^{-4}$ atm and $T=1002$ K, $K_C=\dfrac{2.2\times10^{-4}}{0.0821\times1002}\approx2.67\times10^{-6}$ mol L $^{-1}$ . It equally tells us, purely from the sign of $\Delta n_g$ , whether $K_C$ must be larger or smaller than $K_P$ for a given reaction without any numbers at all -- if a gaseous homogeneous reaction has more moles of product than reactant ( $\Delta n_g>0$ ), then since $K_P=K_C(RT)^{\Delta n_g}$ with $RT>1$ in most common units, $K_P>K_C$ , i.e. $K_C$ is the smaller of the two.

Step 1. For A(g) + B(g) ⇌ C(g) + D(g), the equilibrium constant in terms of partial pressures is $K_p=\dfrac{P_C\,P_D}{P_A\,P_B}$ .

Step 2. For an ideal gas, $PV=nRT$ , so $P=\dfrac{n}{V}RT=[\text{conc}]\,RT$ . Applying this to each species: $P_A=[A]RT$ , $P_B=[B]RT$ , $P_C=[C]RT$ , $P_D=[D]RT$ .

Step 3. Substituting into the $K_p$ expression: $K_p=\dfrac{[C]RT\cdot[D]RT}{[A]RT\cdot[B]RT}=\dfrac{[C][D]}{[A][B]}\times(RT)^{(1+1)-(1+1)}$ .

Step 4. Since $K_c=\dfrac{[C][D]}{[A][B]}$ , this gives $K_p=K_c(RT)^{\Delta n}$ , where $\Delta n$ = (moles of gaseous products) − (moles of gaseous reactants) = $(1+1)-(1+1)=0$ here.

Step 5. Therefore, for this reaction, $K_p=K_c(RT)^0=K_c$ .

✓Final answer

K_p=K_c(RT)^{\Delta n}

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)

moles of gaseous reactants = 1 + 3 = 4 (from N2 and 3H2)

Δn = 2 - 4 = -2

Substituting:

Kp = Kc (RT)^(-2) = Kc / (RT)^2

Rearranging to get Kc/Kp:

Kc/Kp = (RT)^2

✓Final answer
$The correct option is (c) (RT)^2.$
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

This negative value makes physical sense: the forward reaction (SO2 oxidation, as in the Contact process for sulfuric acid manufacture) reduces the total number of gas molecules (3 moles of gas becoming 2 moles of gas), so by Le Chatelier's principle increasing pressure favours the forward (product) direction — consistent with Delta-ng being negative.

✓Final answer
$The correct option is (d) -1 — Delta-ng = (moles of gaseous products) - (moles of gaseous reactants) = 2 - (2+1) = 2 - 3 = -1.$
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

Option (c), "Kp = Kc(RT)^-ve", is just a restatement of the defining relation rather than the numeric consequence the question is asking for, and options (b) and (d) reverse the actual inequality.

✓Final answer
$The correct option is (a) Kp < Kc.$

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Q.Derive mathematically value of kP for for A(g) + B(g) ⇌ C(g) + D(g)

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