The relationship between enthalpy change ($\Delta H$) and internal energy change ($\Delta E$ or $\Delta U$) is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$ where $\Delta n_g$ is the change in the number of moles of gaseous substances ($\Delta n_g = n_{p(g)} - n_{r(g)}$). For $\Delta H = \Delta E$, the term $\Delta n_g RT$ must be zero, which means $\Delta n_g = 0$. Let's calculate $\Delta n_g$ for each given reaction:

1. $\text{H}_2\text{(g)} + \text{Br}_2\text{(g)} \rightarrow 2\text{HBr(g)}$ $\Delta n_g = 2 - (1 + 1) = 0$ Here, $\Delta H = \Delta E$.

2. $\text{C(s)} + 2\text{H}_2\text{O(g)} \rightarrow 2\text{H}_2\text{(g)} + \text{CO}_2\text{(g)}$ $\Delta n_g = (2 + 1) - 2 = 1$ Here, $\Delta H > \Delta E$.

3. $\text{PCl}_5\text{(g)} \rightarrow \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$ $\Delta n_g = (1 + 1) - 1 = 1$ Here, $\Delta H > \Delta E$.

4. $2\text{CO(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)}$ $\Delta n_g = 2 - (2 + 1) = -1$ Here, $\Delta H < \Delta E$.

Thus, only the first reaction has $\Delta H = \Delta E$.