5. As shown in Figure 1, the diagonal $BD$ of rectangle $ABCD$ passes through the origin $O$, and the sides of the rectangle are parallel to the coordinate axes. The graph of the inverse proportion function $y=\frac{k}{x} (k>0)$ intersects $BC$ and $CD$ at points $M$ and $N$, respectively. If point $A(-2, -2)$, and the area of $\triangle OMN$ is $\frac{3}{2}$, then $k=(\quad)$.
(A) 2.5
(B) 2
(C) 1.5
(D) 1
5. B.

From the fact that the diagonal of a rectangle bisects the area of the rectangle, we know

Let $O H=a, O G=b$. Then
$$
a b=4, H M=\frac{k}{a}, G N=\frac{k}{b} \text {. }
$$

Thus, the points $M\left(\frac{k}{a}, a\right), N\left(b, \frac{k}{b}\right)$.
And $S_{\text {CHOG }}-S_{\triangle O G N}-S_{\triangle O H M}-S_{\triangle M C N}$
$$
=S_{\triangle O M N}=\frac{3}{2} \text {, }
$$

which means $4-\frac{1}{2} b \cdot \frac{k}{b}-\frac{1}{2} a \cdot \frac{k}{a}-\frac{1}{2}\left(b-\frac{k}{a}\right)\left(a-\frac{k}{b}\right)=\frac{3}{2}$.
Given $k>0$, solving yields $k=2$.