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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 $\ce{2H -> H2}$?

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    $\begingroup$ We do, but not in your case, as we are not interested in the reverse reaction.. $\endgroup$
    – Poutnik
    May 25 at 12:45
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    $\begingroup$ There is no point in doing so here, just make sure the expression for the resulting enthalpy holds true algebraically for reactants and products. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Don't use ALL CAPS, it's rude and hearts readability. $\endgroup$
    – andselisk
    May 25 at 12:50

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