Suppose we have a time-continuous Markov-Process $(X_t)_{t\in\mathbb{R}}$. If $X_0=j$, then $X_t$ needs a random amount of time, say $T_1$, before the process jumps to a new state $i$. $T_1$ is a continuous random variable. All the random variables $T_1, T_2, T_3,\dots$ between the states are exponentially distributed, i.e. $T_k\sim\text{Exp}(\lambda_k)$, for $\lambda>0$ and $k\in\mathbb{N}$. The rate of jumping from from a $j\in X_t$ to $i$ is given by a certain density $\sigma_t(j\to i)$.
Now a asked myself if it was possible to apply the central limit theorem on this certain situation? And if yes, how? Thank you very much for your help!
In the case where the CTMC is finite and time-homogeneous, i.e., $\sigma_t$ is independent of $t$ and $\lambda$ only depends on the current state (which I'll denote al $\lambda_j$ for state $j$), you can apply the CLT to the average waiting time in each state.
Formally, let $(i_n)_{n\in\mathbb{N}}$ be the sequence of indices s.t. we are in state $j$ after the $i_n$-th transition. Then $S_{j,n}=\sum_{k=0}^m \frac{T_{i_k}}{n}$ gives the average time for a transition from $j$, and by the CLT this approaches a normal distribution with expected value $1/\lambda_j$ and variance $1/(\lambda_j^2n)$, i.e., $S_{j,n}\approx \mathcal{N}(\frac{1}{\lambda_j},\frac{1}{\lambda_j^2n})$ for large $n$.
Moreover, you can compute the expected number of visits for each state by viewing your process as a discrete time Markov chain. Let $E_n(j)$ denote the number of times we expect to visit state $j$ within the first $n$ transitions (regardless of how much time they take). Let $P(j,k)$ denote the probability of being in state $j$ after $k$ transitions. Clearly $E_n(j)=\sum_{k=0}^nP(j,k)$. In the limit, $P(j,n)$ approaches the steady-state probabilities when viewing the system as a discrete time Morkov chain. So if we denote the steady-state probability of state $j$ as $E(j)$ we have $E_n(j)\approx nE(j)$ for large $n$.
Then the expected time of a run of length $n$ is just the expected number of times we visit state $j$, times the expected waiting time in state $j$, summed over all states:
$$\sum_{j\in S}nE(j)\mathcal{N}(\frac{1}{\lambda_j},\frac{1}{\lambda_j^2n})$$
for large $n$, which is just a mixture of normal distributions.