$\Delta n_g = 2 - (1+3) = 2 - 4 = -2$
$$K_p = K_c(RT)^{\Delta n_g} = K_c(RT)^{-2} = \frac{K_c}{(RT)^2}$$At 300 K: $RT = 0.0821 \times 300 = 24.63$. So $(RT)^2 = 606$.
$K_p = K_c/606 \ll K_c$
Statement I is TRUE. Statement II gives the correct explanation. Answer: Both correct, II explains I
For a general gaseous reaction $aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)$: $K_p = K_c(RT)^{\Delta n_g}$ where $\Delta n_g = (c+d) - (a+b)$ = moles of gaseous products - moles of gaseous reactants. $R = 0.0821$ L·atm/(mol·K) or $R = 8.314$ J/(mol·K). Three cases: $\Delta n_g = 0$: $K_p = K_c$. Examples: H₂+I₂⇌2HI, N₂+O₂⇌2NO. $\Delta n_g > 0$: $K_p > K_c$ (e.g., PCl₅⇌PCl₃+Cl₂, $\Delta n_g=1$; N₂O₄⇌2NO₂, $\Delta n_g=1$; CaCO₃⇌CaO+CO₂, $\Delta n_g=1$). $\Delta n_g < 0$: $K_p < K_c$ (e.g., N₂+3H₂⇌2NH₃, $\Delta n_g=-2$; 2SO₂+O₂⇌2SO₃, $\Delta n_g=-1$). Derivation: using ideal gas law $P = nRT/V$, concentration $[A] = P_A/(RT)$. Substituting into $K_p$ expression and rearranging gives $K_p = K_c(RT)^{\Delta n_g}$.
Strictly speaking, equilibrium constants are dimensionless. This is achieved by dividing each concentration by standard concentration $c^\circ = 1$ mol/L or each pressure by standard pressure $P^\circ = 1$ bar or 1 atm. $K_c = \prod_i ([A_i]/c^\circ)^{\nu_i}$ (numerically same as using concentrations in mol/L). $K_p = \prod_i (P_i/P^\circ)^{\nu_i}$ (dimensionless). In practice: use concentrations in mol/L for $K_c$ and pressures in atm or bar for $K_p$. The numerical value of $K_p$ depends slightly on the choice of pressure standard (1 atm vs 1 bar). The relationship $K_p = K_c(RT)^{\Delta n_g}$ uses $R = 0.0821$ L·atm/(mol·K) when pressure is in atm and concentration in mol/L.
Understanding $K_p$ vs $K_c$ has practical implications: Haber process ($N_2+3H_2 \to 2NH_3$, $\Delta n_g = -2$): $K_p << K_c$. At high pressures, $K_p$ is calculated but partial pressures are relevant for yield calculation. Increasing total pressure at constant temperature: doesn't change $K_p$ (constant at constant T), but shifts reaction forward (Le Chatelier: fewer moles of gas). Contact process ($2SO_2+O_2 \to 2SO_3$, $\Delta n_g=-1$): $K_p = K_c/(RT)$. High pressure also favours SO₃ production. Dissociation of N₂O₄ ($N_2O_4 \to 2NO_2$, $\Delta n_g=+1$): $K_p > K_c$. Decreasing pressure shifts forward (more moles of gas).
For $A(g) \rightleftharpoons nB(g)$ starting with 1 mole A at pressure P: At equilibrium: A = $1-\alpha$, B = $n\alpha$, total = $1+(n-1)\alpha$. Mole fractions: $x_A = (1-\alpha)/(1+(n-1)\alpha)$, $x_B = n\alpha/(1+(n-1)\alpha)$. Partial pressures: $P_A = x_A P$, $P_B = x_B P$. $K_p = P_B^n/P_A = (n\alpha P/(1+(n-1)\alpha))^n / ((1-\alpha)P/(1+(n-1)\alpha))$. For PCl₅ ⇌ PCl₃ + Cl₂ ($n=2$ products): $K_p = \alpha^2 P/(1-\alpha^2)$. At higher P: $K_p$ unchanged (constant T) but $\alpha$ decreases (Le Chatelier: more pressure → less dissociation). At lower P: $\alpha$ increases. For N₂O₄ ⇌ 2NO₂: $K_p = 4\alpha^2 P/(1-\alpha^2)$. Brown colour of NO₂ intensifies at low pressure (more dissociation).
$\Delta G^\circ = -RT\ln K = -2.303RT\log K$. At 298 K: $\Delta G^\circ = -5.7\log K$ kJ/mol. If $K > 1$: $\log K > 0$, $\Delta G^\circ < 0$ (products favoured). If $K < 1$: $\Delta G^\circ > 0$ (reactants favoured). If $K = 1$: $\Delta G^\circ = 0$ (equal products and reactants). $\Delta G = \Delta G^\circ + RT\ln Q$. At equilibrium: $\Delta G = 0$, $Q = K$. Van't Hoff equation: $\frac{d\ln K}{dT} = \frac{\Delta H^\circ}{RT^2}$. Integrating: $\ln\frac{K_2}{K_1} = \frac{\Delta H^\circ}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$. If $\Delta H^\circ > 0$: $K$ increases with $T$. If $\Delta H^\circ < 0$: $K$ decreases with $T$. This is Le Chatelier's principle in quantitative form.
Heterogeneous equilibria involve species in more than one phase. Pure solids and pure liquids are NOT included in the equilibrium constant expression (their activities = 1, concentration is constant within the pure phase). $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$: $K_p = P_{CO_2}$, $K_c = [CO_2]$. The equilibrium depends only on $P_{CO_2}$, not on amounts of CaCO₃ or CaO present. $C(s) + H_2O(g) \rightleftharpoons CO(g) + H_2(g)$: $K_p = P_{CO} \cdot P_{H_2}/P_{H_2O}$. $FeO(s) + CO(g) \rightleftharpoons Fe(s) + CO_2(g)$: $K_p = P_{CO_2}/P_{CO}$. For dissolution: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$: $K = K_{sp} = [Ag^+][Cl^-]$ (water activity = 1).
When reactions are combined, equilibrium constants multiply. If: Reaction 1: A ⇌ B, $K_1$. Reaction 2: B ⇌ C, $K_2$. Then overall: A ⇌ C, $K = K_1 K_2$. If Reaction 1 reversed: B ⇌ A, $K = 1/K_1$. If Reaction 1 multiplied by $n$: $nA ⇌ nB$, $K = K_1^n$. Applications: Hess law for $K$ values. Stepwise formation constants of complexes: $\beta_n = \beta_1 \times \beta_2 \times ... \times \beta_n$. For acid-base equilibria: $K_a \times K_b = K_w$ (for conjugate acid-base pair). This is why stronger acid has weaker conjugate base.
Adding an ion that is already involved in the equilibrium shifts the equilibrium according to Le Chatelier. Common ion effect reduces solubility: AgCl(s) ⇌ Ag⁺+Cl⁻. Adding NaCl → [Cl⁻] increases → Q > K → equilibrium shifts back → less AgCl dissolves. Common ion effect in buffer solutions: CH₃COOH ⇌ CH₃COO⁻ + H⁺. Adding CH₃COONa (common CH₃COO⁻) → [CH₃COO⁻] increases → Q > K → shift back (fewer H⁺ → higher pH). This is how buffer works. Common ion effect in precipitation: adding excess precipitating ion to remove a specific cation completely from solution. In qualitative analysis: adding excess Cl⁻ in Group I ensures complete precipitation of Ag⁺, Pb²⁺, Hg₂²⁺.