5. If for any real numbers $x \neq y$, we have
$$
\frac{\mathrm{e}^{x}-\mathrm{e}^{y}}{x-y}+k(x+y)>1,
$$

then the range of the real number $k$ is
5. $\left\{-\frac{1}{2}\right\}$.

The problem is equivalent to $f(x)=\mathrm{e}^{x}+k x^{2}-x$ being monotonically increasing on $\mathbf{R}$, i.e.,
$$
f^{\prime}(x)=\mathrm{e}^{x}+2 k x-1 \geqslant 0
$$

always holds.
When $x=0$, $f^{\prime}(x)=0$, then $f^{\prime}(x)$ attains its minimum value outside $x=0$, so $f^{\prime \prime}(0)=0$.
$$
\begin{array}{l}
\text { Also, } f^{\prime \prime}(x)=\mathrm{e}^{x}+2 k \text {, hence } \\
1+2 k=0 \Rightarrow k=-\frac{1}{2} .
\end{array}
$$

Verify that it satisfies the condition.