JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Equilibrium

The link between the two constants is $K_p=K_c(RT{)}^{\Delta n_g}$, so $K_p/K_c=(RT{)}^{\Delta n_g}$ and only the change in moles of gas matters. For $\mathrm{N}{}_2+\mathrm{O}{}_2\to 2\mathrm{NO}$ the gas moles are unchanged, $\Delta n_g=0$, so the ratio is $1$. For $\mathrm{N}{}_2\mathrm{O}{}_4\to 2\mathrm{NO}{}_2$ one mole becomes two, $\Delta n_g=+1$, so the ratio is $RT=24.62$ dm${}^3$ atm mol${}^{-1}$. For $\mathrm{N}{}_2+3\mathrm{H}{}_2\to 2\mathrm{NH}{}_3$ four moles become two, $\Delta n_g=-2$, so the ratio is $(RT{)}^{-2}=1/24.62^2\approx 1.65\times 10^{-3}$ dm${}^{-6}$ atm${}^{-2}$ mol${}^2$. Read off in turn these are $1$, then $24.62$, then $1.65\times 10^{-3}$.