# Is Kp and Kc only effected by temperature?

Are Kp and Kc only effected by temperature?

So in exothermic and endothermic equilibrium reactions you are increasing or decreasing temperature, which will effect which side is favourable.

But then so does pressure if one side has fewer molecules?

• – Philipp Jun 3 '17 at 12:58
• @Philipp is this statement in my textbook wrong then "increasing the pressure forces the equilibrium in such a way as to reduce the total pressure, that is, to the side with fewer molecules in other words KP increases. So increasing the total pressure increases the yield of ammonia" – Aaron Jun 3 '17 at 13:23
• It sounds fishy, at least the part "in other words KP increases". They are talking about Le Chatelier's principle but that does not entail a pressure dependence of the equilibrium constant. However, the partial pressures/activities/relative amounts of the reactants and products in the reaction mixture which feature in the expression for $K$ do not need to be constant with pressure only their reaction quotient has to be in order to preserve the contant $K$. Thus, pressure (depending on how it is administered) does influence the equilibrium composition of the reaction mixture. – Philipp Jun 3 '17 at 15:14
• You could have a look at page 221 of the 9th edition of the book "Physical Chemistry" by Atkins. It describes this in more detail. – Philipp Jun 3 '17 at 15:18
• As $K_p$ is defined by standard states, i.e. 1 atm (or 1 bar) it must be independent of pressure. The T dependence is from the van't Hoff isochore $d\ln(K_p)/dT = \Delta H^o/RT^2$ (assuming $\Delta H$ constant). If the equilibrium constant is defined in terms of mole fraction x then $K_x=K_pp^{-\Delta n}$ where $\Delta n =a+b-c-d$ is the change in number of moles in reaction $\ce{aA + bB <=>cC + dD}$. Sometimes the equilibrium constant is called $K_p$ when it should be $K_x$ or $K_c=K_p(RT)^{-\Delta n}$. – porphyrin Jun 3 '17 at 16:08