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
$$\ce{C(s) + 4 H(g) -> CH4(g)}\tag{R3}$$
Answer
$\Delta_\mathrm{r}H^\circ = \pu{-950.4 kJ mol^-1}$.
Question
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 $\ce{2H -> H2}$?
