Capacitive reactance related to frequency
The more we increase the capacitance of a capacitor -&gt; for the same charge at the plates of the capacitor we get less voltage which resists current from the AC source.

First, let's look at how the capacitive reactance is obtained.
The relationship between electrical charge and current is:
$$
dq = i\ dt
$$
where \$q\$ is the electrical charge, \$i\$ is the current and \$t\$ is the time.
The change of electrical charge stored by the capacitor is:
$$
dq = C\ dV
$$
where \$C\$ is the capacitance and \$V\$ is the voltage across the capacitor.
From these two equations, we obtain the famous capacitor current-voltage relationship:
$$
i\ dt = C\ dV \\
\Rightarrow i_C = C \ \frac{dV_C}{dt}
$$
We can replace \$d/dt\$ with \$j\omega\$ (or \$2\pi f\$), so finally we obtain:
$$
i_C = V_C \ j\omega\ C \\
\Rightarrow V_C = i_C \ \frac{1}{j\omega \ C}
$$
The impedance of the capacitor is
$$
Z_C=\frac{i_C}{V_C}=\frac{1}{j\omega \ C}=\frac{1}{j\ 2\pi f \ C}
$$
and the capacitive reactance is the magnitude of the impedance:
$$
X_C=|Z_C|=\frac{1}{2\pi f \ C}
$$
So,

the higher the frequency, the lower the capacitive reactance
the higher the capacitance, the lower the capacitive reactance