# Is there a need to change sign for enthalpy value for a reaction that appears in reverse when using Hess's law?

### Problem

Given the enthalpies of the following reactions

\begin{align} \ce{2H(g) &-> H2(g)} &\quad \Delta_\mathrm{r}H^\circ &= \pu{-437.6 kJ mol^-1} \tag{R1}\\ \ce{C(s) + 2 H2(g) &-> CH4(g)} &\quad \Delta_\mathrm{r}H^\circ &= \pu{-75.2 kJ mol^-1} \tag{R2} \end{align}

find $$\Delta_\mathrm{r}H^\circ$$ of

$$\ce{C(s) + 4 H(g) -> CH4(g)}\tag{R3}$$

$$\Delta_\mathrm{r}H^\circ = \pu{-950.4 kJ mol^-1}$$.
Why don't we change the sign of $$\pu{-437.6 kJ mol^-1}$$ (times two) if we go from $$\ce{2 H2}$$ to $$\ce{4H}?$$ Isn't it endothermic when we reverse the reaction $$\ce{2H -> H2}$$?