# Kc(equilibrium constant) and Kp

Its given in the book that at 12.18kelvin Kc=Kp. How is that equated from the equation Kp= Kc(RT)^∆ng. I am unable to sort it out, please guide me in to the topic....

• It seems like a homework question. Can you please give some more clarification? Jul 19 '16 at 6:23
• Hi; you should define your variables more clearly and then use the equation formatting tools to make nice equations. The easiest way to do this is to click on some other questions with equations, and copy the syntax. Also there is some information at the header of the box you use to post equations. Jul 19 '16 at 9:38

From the ideal gas equation, we get $$PV=nRT \implies P=\frac{n}{V}RT=cRT$$ where $c$ represents the concentration.
Consider an arbitrary reaction $$n_AA\rightarrow n_BB$$
$$K_c=\frac{c_B^{n_B}}{c_A^{n_A}}$$ $$K_p=\frac{P_B^{n_B}}{P_A^{n_A}}=\frac{(c_BRT)^{n_B}}{(c_ART)^{n_A}}=K_c \times (RT)^{\Delta n}$$
Most probably, the reaction you are talking about has $\Delta n=0$.
Actually you can see here that in formula $$K_p=K_c×(RT)^{\partial n}$$; $$K_p$$ will be equal to $$K_c$$ if and only if $$\partial n=0$$ or $$RT = 1$$ So when $$RT =1$$ then $$T =\frac1R$$ that is $$\frac{1}{0.0821}=\pu{12.18K}$$.