For the gas-phase reaction $2\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{SO_3(g)}$, the value of $\Delta n$ (used in $K_p = K_c (RT)^{\Delta n}$) is

1. $\Delta n = $ (moles of GASEOUS products) − (moles of GASEOUS reactants). 2. Products: 2 mol $\mathrm{SO_3(g)}$. 3. Reactants: 2 mol $\mathrm{SO_2(g)} + 1$ mol $\mathrm{O_2(g)} = 3$ mol. 4. $\Delta n = 2 - 3 = -1$. 5. Therefore $K_p = K_c (RT)^{-1}$, i.e. $K_p < K_c$ at usual temperatures (since $RT > 1$ when $R$ is in $\text{L\,bar/(K\,mol)}$ and $T$ in K). 6. Option A flips the sign. Option B would apply if products and reactants had equal gas-mole sums. Option C would correspond to $4$ products and $2$ reactants. _Source: NCERT Class 11 Chemistry, Ch 6, §6.4.1 (Eq. 6.15 with $\Delta n$ definition), p. 11._