5. As shown in Figure 1, in the tetrahedron $A B C D$, it is known that $A B \perp$ plane $B C D, \triangle B C D$ is an equilateral triangle with a side length of 3. If $A B$ $=2$, then the surface area of the circumscribed sphere of the tetrahedron $A B C D$ is $\qquad$
5. $16 \pi$.

As shown in Figure 4, let the center of the equilateral $\triangle BCD$ be $O_{1}$, and the center of the circumscribed sphere of the tetrahedron $ABCD$ be $O$.
Then $O O_{1} \perp$ plane $BCD$, $O O_{1} / / AB$.
Thus, $B O_{1}=\frac{2}{3} \times \frac{\sqrt{3}}{2} \times 3=\sqrt{3}$.
Take the midpoint $E$ of side $AB$.
Since $O A=O B$, we know $O E \perp AB$, $O E / / O_{1} B$.
Therefore, $O O_{1}=E B=1$.
Hence, $O A=O B=2$.
Therefore, the radius of the circumscribed sphere of the tetrahedron $ABCD$ is 2, and its surface area is $16 \pi$.