TLDR; Because only the concentration equilibrium constant is directly related to the rate constants themselves.

Given that pressure has a minimal effect on equilibria in solution, I'm going to answer assuming you're interested in gas-phase equilibria. The first step in showing why pressure/volume matter to $K_x$ is to move to the partial-pressure equilibrium constant, $K_p$. For a system composed entirely of ideal gases, the partial pressure of each species is related to its concentration as:

$$ {P_i\over RT} = {n_i\over V} = C_i $$

Substituting this into the expression for $K_c$ for your example reaction gives:

$$ K_c = {C_P C_Q\over C_A} = {1\over RT}{P_P P_Q\over P_A} = {1\over RT}K_p $$

Thus, for this reaction:

$$ K_p = RT\,K_c \tag{1}\label{Kp} $$

For other reactions, the power of $RT$ on the RHS of Eq. $\eqref{Kp}$ can differ, depending on the change in total moles between reactants and products.

Since $K_c = f(T)$ and $K_p = (RT)^nK_c$, then $K_p$ is also dependent only on temperature (with ideal gases only!), not pressure or volume.

Where pressure/volume enter the picture is when one goes from $K_p$ to $K_x$. Again assuming ideal behavior, the mole fraction of a species is defined as:

$$ P_i = x_i P $$

Substituting this into the $K_P$ expression:

$$ K_p = {P_P P_Q\over P_A} = P {x_P x_Q\over x_A} = PK_x $$

Thus, for this reaction:

$$ K_x = {1\over P}K_p \tag{2}\label{Kx} $$

Similar to the above, the power of $1\over P$ in Eq. $\eqref{Kx}$ can vary, depending on the change in total moles between reactants and products.

In any event, this is where the dependence on $P$ enters the expression for $K_x$.

This dependence on pressure implies a related dependence on volume, since by the ideal gas law:

$$ V = {nRT \over P} $$

and, equivalently:

$$ P = {nRT \over V}\tag{3}\label{igl-P} $$

If one is working with a system where the volume and temperature are the control variables but the pressure is not, then in general the pressure will vary implicitly according to Eq. $\eqref{igl-P}$. If the temperature and/or volume change in any way other than proportionally $($i.e., ${T\over V} = \text{constant})$, then the pressure dependence of Eq. $\eqref{Kx}$ will manifest itself in a fashion that looks like a volume dependence.

TLDR; Because only the concentration equilibrium constant is directly related to the rate constants themselves.

Given that pressure has a minimal effect on equilibria in solution, I'm going to answer assuming you're interested in gas-phase equilibria. The first step in showing why pressure/volume matter to $K_x$ is to move to the partial-pressure equilibrium constant, $K_p$. For a system composed entirely of ideal gases, the partial pressure of each species is related to its concentration as:

$$ {P_i\over RT} = {n_i\over V} = C_i $$

Substituting this into the expression for $K_c$ for your example reaction gives:

$$ K_c = {C_P C_Q\over C_A} = {1\over RT}{P_P P_Q\over P_A} = {1\over RT}K_p $$

Thus, for this reaction:

$$ K_p = RT\,K_c \tag{1}\label{Kp} $$

For other reactions, the power of $RT$ on the RHS of Eq. $\eqref{Kp}$ can differ, depending on the change in the sum of the stoichiometric coefficients between the reactants and the products.

Since $K_c = f(T)$ and $K_p = (RT)^nK_c$, then $K_p$ is also dependent only on temperature (with ideal gases only!), not pressure or volume.

Where pressure/volume enter the picture is when one goes from $K_p$ to $K_x$. Again assuming ideal behavior, the mole fraction of a species is defined as:

$$ P_i = x_i P $$

Substituting this into the $K_P$ expression:

$$ K_p = {P_P P_Q\over P_A} = P {x_P x_Q\over x_A} = PK_x $$

Thus, for this reaction:

$$ K_x = {1\over P}K_p \tag{2}\label{Kx} $$

Similar to the above, the power of $1\over P$ in Eq. $\eqref{Kx}$ can vary, depending on the change in the sums of the stoichiometric coefficients between the reactants and the products.

In any event, this is where the dependence on $P$ enters the expression for $K_x$.

This dependence on pressure implies a related dependence on volume, since by the ideal gas law:

$$ V = {nRT \over P} $$

and, equivalently:

$$ P = {nRT \over V}\tag{3}\label{igl-P} $$

If one is working with a system where the volume and temperature are the control variables but the pressure is not, then in general the pressure will vary implicitly according to Eq. $\eqref{igl-P}$. If the temperature and/or volume change in any way other than proportionally $($i.e., ${T\over V} = \text{constant})$, then the pressure dependence of Eq. $\eqref{Kx}$ will manifest itself in a fashion that looks like a volume dependence.

TLDR; Because only the concentration equilibrium constant is directly related to the rate constants themselves.

Given that pressure has a minimal effect on equilibria in solution, I'm going to answer assuming you're interested in gas-phase equilibria. The first step in showing why pressure/volume matter to $K_x$ is to move to the partial-pressure equilibrium constant, $K_p$. For a system composed entirely of ideal gases, the partial pressure of each species is related to its concentration as:

$$ {P_i\over RT} = {n_i\over V} = C_i $$

Substituting this into the expression for $K_c$ for your example reaction gives:

$$ K_c = {C_P C_Q\over C_A} = {1\over RT}{P_P P_Q\over P_A} = {1\over RT}K_p $$

Thus, for this reaction:

$$ K_p = RT\,K_c \tag{1}\label{Kp} $$

For other reactions, the power of $RT$ on the RHS of Eq. $\eqref{Kp}$ can differ, depending on the change in total number of moles between reactants and products.

Since $K_c = f(T)$ and $K_p = (RT)^nK_c$, then $K_p$ is also dependent only on temperature (with ideal gases only!), not pressure or volume.

Where pressure/volume enter the picture is when one goes from $K_p$ to $K_x$. Again assuming ideal behavior, the mole fraction of a species is defined as:

$$ P_i = x_i P $$

Substituting this into the $K_P$ expression:

$$ K_p = {P_P P_Q\over P_A} = P {x_P x_Q\over x_A} = PK_x $$

Thus, for this reaction:

$$ K_x = {1\over P}K_p \tag{2}\label{Kx} $$

Similar to the above, the power of $1\over P$ in Eq. $\eqref{Kx}$ can vary, depending on the change in total number of moles between reactants and products.

In any event, this is where the dependence on $P$ enters the expression for $K_x$.

This dependence on pressure implies a related dependence on volume, since by the ideal gas law:

$$ V = {nRT \over P} $$

and, equivalently:

$$ P = {nRT \over V}\tag{3}\label{igl-P} $$

If one is working with a system where the volume and temperature are the control variables but the pressure is not, then in general the pressure will vary implicitly according to Eq. $\eqref{igl-P}$. If the temperature and/or volume change in any way other than proportionally $($i.e., ${T\over V} = \text{constant})$, then the pressure dependence of Eq. $\eqref{Kx}$ will manifest itself in a fashion that looks like a volume dependence.

TLDR; Because only the concentration equilibrium constant is directly related to the rate constants themselves.

Given that pressure has a minimal effect on equilibria in solution, I'm going to answer assuming you're interested in gas-phase equilibria. The first step in showing why pressure/volume matter to $K_x$ is to move to the partial-pressure equilibrium constant, $K_p$. For a system composed entirely of ideal gases, the partial pressure of each species is related to its concentration as:

$$ {P_i\over RT} = {n_i\over V} = C_i $$

Substituting this into the expression for $K_c$ for your example reaction gives:

$$ K_c = {C_P C_Q\over C_A} = {1\over RT}{P_P P_Q\over P_A} = {1\over RT}K_p $$

Thus, for this reaction:

$$ K_p = RT\,K_c \tag{1}\label{Kp} $$

For other reactions, the power of $RT$ on the RHS of Eq. $\eqref{Kp}$ can differ, depending on the change in total moles between reactants and products.

Since $K_c = f(T)$ and $K_p = (RT)^nK_c$, then $K_p$ is also dependent only on temperature (with ideal gases only!), not pressure or volume.

Where pressure/volume enter the picture is when one goes from $K_p$ to $K_x$. Again assuming ideal behavior, the mole fraction of a species is defined as:

$$ P_i = x_i P $$

Substituting this into the $K_P$ expression:

$$ K_p = {P_P P_Q\over P_A} = P {x_P x_Q\over x_A} = PK_x $$

Thus, for this reaction:

$$ K_x = {1\over P}K_p \tag{2}\label{Kx} $$

Similar to the above, the power of $1\over P$ in Eq. $\eqref{Kx}$ can vary, depending on the change in total moles between reactants and products.

In any event, this is where the dependence on $P$ enters the expression for $K_x$.

This dependence on pressure implies a related dependence on volume, since by the ideal gas law:

$$ V = {nRT \over P} $$

and, equivalently:

$$ P = {nRT \over V}\tag{3}\label{igl-P} $$

If one is working with a system where the volume and temperature are the control variables but the pressure is not, then in general the pressure will vary implicitly according to Eq. $\eqref{igl-P}$. If the temperature and/or volume change in any way other than proportionally $($i.e., ${T\over V} = \text{constant})$, then the pressure dependence of Eq. $\eqref{Kx}$ will manifest itself in a fashion that looks like a volume dependence.