Sign of change in enthalpy and change in entropy
TL;DR
In general the entropy of reaction can be written as
$$  T\Delta _r S=  \Delta_r H  + RT\log \left(\frac{Q_e}{Q}\right)  $$
At equilibrium $Q_e=Q$ and
$$  T\Delta _r S_e=  \Delta_r H   $$

Consider a simple reaction that behaves ideally (occurs under ideal solution conditions).
If it is carried out at constant T and p we can write
$$\Delta_r G = \Delta_r G^\circ + RT \log Q \tag{1}$$
where Q is the reaction quotient.
But we can also write that
$$\Delta_r G = \Delta_r H - T\Delta_r S\tag{2a}$$
and
$$\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ\tag{2b}$$
Equation (1) can then be written as
$$\Delta_r G = \Delta_r H^\circ - T(\Delta_r S^\circ-R \log Q) \tag{3}$$
Matching terms in equations (2a) and (3) we have that
$$\Delta_r H = \Delta_r H^\circ \tag{4a}$$
and
$$\Delta_r S = \Delta_r S^\circ - R\log Q \tag{4b}$$
When the reaction is at equilibrium $Q=Q_e$ (the reaction quotient is then equal to the equilibrium constant, here written $Q_e$) and $\Delta_r G = 0$ which means, combining equations (2a) and (4a) that
$$ T\Delta _r S_e =  \Delta _r H^ \circ \tag{5} $$
and
$$ T\Delta _r S^ \circ = T\Delta_r S_e + RT\log Q_e  \tag{6}$$
so that
$$  T\Delta _r S=  T\Delta_r S_e + RT\log Q_e  - RT\log Q \tag{7a} $$
or
$$  T\Delta _r S=  \Delta_r H ^\circ + RT\log\left(\frac{Q_e}{Q}\right)  \tag{7b} $$
Now compare equations (5) and (7b). Equation (5) holds at equilibrium and says, sure enough, that the reaction entropy and enthalpy are equal in sign at this point in the reaction coordinate. However, equation (7b) - which is the more general expression - says that $\Delta_r S$ can in fact differ in sign from $\Delta_r H^\circ$, depending on the magnitude of the reaction quotient Q. It turns out that while the enthalpy of a reaction in an ideal solution is a constant, the entropy of reaction can be tuned by modifying Q.