(a) $K_{p} = K_{c}(RT)^{\mathrm{\Delta}n_{g}}$ $\mathrm{\Delta}n_{g} = 1 - 1.5 = - 0.5$ $K_{p} = K_{c}(RT)^{\frac{- 1}{2}} = \frac{K_{c}}{(RT)^{\frac{1}{2}}}$ $\therefore$ $\frac{K_{c}}{K_{p}} = (RT)^{\frac{1}{2}}$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.