The ratio of rate constants for the forward and backward reactions gives us the equilibrium constant, $K_c$:

$$ \ce{A \underset{k_{-1}}{\overset{k_1}{<=>}} P + Q}\quad \Longrightarrow \quad K_c = {k_1\over k_{-1}} = {C_P C_Q\over C_A} $$

Per kinetic theory, rate constants depend only on temperature and the presence of any catalysts:

$$ k_i = A_i e^{-E_{a,i}\over RT} $$

Thus, their ratio (i.e., the equilibrium constant, $K_c$) also must depend only on temperature:

So: Why does $K_x$, the mole-fraction equilibrium constant, depend on pressure and volume?

The ratio of rate constants for the forward and backward reactions gives us the equilibrium constant, $K_c$. For a sample reaction:

$$ \ce{A \underset{k_{-1}}{\overset{k_1}{<=>}} P + Q}\quad \Longrightarrow \quad K_c = {k_1\over k_{-1}} = {C_P C_Q\over C_A} $$

Per kinetic theory, rate constants depend only on temperature and the presence of any catalysts:

$$ k_i = A_i e^{-E_{a,i}\over RT} $$

Thus, their ratio (i.e., the equilibrium constant, $K_c$) also must depend only on temperature:

So: Why does $K_x$, the mole-fraction equilibrium constant, depend on pressure and volume?

The ratio of rate constants for forward and backward reaction gives us the equilibrium constant (say kc).

1.Rate constants depend on temperature and catalyst (Kinetics). 2.Their ratio i.e equilibrium constant depends only on temperature

3.So, why does K(x) (mole fraction equilibrium constant) depend on pressure and volume?

Also I need to ask why equilibrium constants are unaffected by catalyst? Does the fact (energy level diagram) that catalyst increase rate of reaction for both ways equally accounts the independence of equilibrium constants on catalyst.

Also is there some sense in saying that temperature being an intensive property persist on taking ratio as well..?

The ratio of rate constants for the forward and backward reactions gives us the equilibrium constant, $K_c$:

$$ \ce{A \underset{k_{-1}}{\overset{k_1}{<=>}} P + Q}\quad \Longrightarrow \quad K_c = {k_1\over k_{-1}} = {C_P C_Q\over C_A} $$

Per kinetic theory, rate constants depend only on temperature and the presence of any catalysts:

$$ k_i = A_i e^{-E_{a,i}\over RT} $$

Thus, their ratio (i.e., the equilibrium constant, $K_c$) also must depend only on temperature:

So: Why does $K_x$, the mole-fraction equilibrium constant, depend on pressure and volume?

The ratio of rate constants for forward and backward reaction gives us the equilibrium constant (say kc).

1.Rate constants depend on temperature and catalyst (Kinetics). 2.Their ratio i.e equilibrium constant depends only on temperature

3.So, why does K(x) (mole fraction equilibrium constant) depend on pressure and volume.**

ALSO I NEED TO ASK WHY EQUILIBRIUM CONSTANTS ARE UNAFFECTED BY CATALYST Does the fact (ENERGY LEVEL DIAGRAM)that catalyst increase rate of reaction for both ways equally accounts the independence of equilibrium constants on catalyst.

Also is there some sense in saying that temperature being an intensive property persist on taking ratio as well..?

The ratio of rate constants for forward and backward reaction gives us the equilibrium constant (say kc).

1.Rate constants depend on temperature and catalyst (Kinetics). 2.Their ratio i.e equilibrium constant depends only on temperature

3.So, why does K(x) (mole fraction equilibrium constant) depend on pressure and volume?

Also I need to ask why equilibrium constants are unaffected by catalyst? Does the fact (energy level diagram) that catalyst increase rate of reaction for both ways equally accounts the independence of equilibrium constants on catalyst.

Also is there some sense in saying that temperature being an intensive property persist on taking ratio as well..?