Why does activation energy depend on temperature?
Although the question has been already answered, I will like to add a bit more of detail.
The expression you are quoting come from the thermodynamic formulation of the TST that expresses the dependence of the rate constant of temperature as
$$k(T)=\frac{k_\mathrm bT}h\exp\left(-\frac{-\Delta G^\ddagger_0}{RT}\right)$$
and can be partitioned in an entropic and enthalpic contributions
$$k(T)=\frac{k_\mathrm bT}h\exp\left(-\frac{-\Delta S^\ddagger_0}R\right)\exp \left(-\frac{-\Delta H^\ddagger_0}{RT}\right)$$
From this equation you can obtain the activation energy $E_\mathrm a$ using the relation
$$\frac{\mathrm d\ln k(t)}{\mathrm dT}=-\frac{E_\mathrm a}{RT^2}$$
Using this equation you will obtain
$$E_\mathrm a=\Delta U^\ddagger_0+RT$$
Since $H=U+pV$
$$\Delta H^\ddagger_0=\Delta U^\ddagger_0+p\,\Delta V^\ddagger_0+RT$$
For a condensed phase reaction $\Delta V^\ddagger_0\approx0$ and
$$E_\mathrm a=\Delta H^\ddagger_0+RT$$
For a gas phase reaction, $p\Delta V^\ddagger_0=\Delta n^\ddagger RT$.
For a gas phase unimolecular reaction $\Delta n^\ddagger=0$, the result is equal to that the obtained by a condensed phase reaction, while for a second-order reaction $\Delta n^\ddagger=-1$, and
$$E_\mathrm a=\Delta H^\ddagger_0+2RT$$
You can find a more detailed discussion in Keith Laidler, Chemical kinetics 3rd. ed., Harper Collins Publishers, (1987); Jeffrey I. Steinfeld, Joseph S. Francisco, William L. Hase, Chemical Kinetics and Dynamics, Prentice Hall (1989); or Paul L. Houston, Chemical Kinetics and Reaction Dynamics, McGraw-Hill (2001).