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The $\ce{H-H}$ bond enthalpy is $436 \mathrm{kJ/mol}$:
$$\ce{H2 (g) -> 2H (g)} \qquad \Delta H = +436 \mathrm{kJ/mol}$$
Does this only apply at the standard temperature of $298 \text{K}$, i.e. would the bond enthalpy vary if I decrease or increase the temperature?
And is it practically just so that, for a temperature $T$, as long as you pump the bond enthalpy’s worth of kilojoules into the substance, it will be atomized?
The bond dissociation enthalpy, like any enthalpy of reaction, will have some variation with temperature. This change in enthalpy can be determined using Kirchoff's Law: $$H_{T_f}=H_{T_i}+\int_{T_{f}}^{T_{i}}\Delta c_p(T')dT'$$ Where $$\Delta c_p(T')=\sum_i\nu_ic_{p,i}(T')$$ so that $c_{p,i}$ is the constant pressure heat capacity of the $\mathrm{i^{th}}$ compound in a reaction and $\nu_i$ is the the reaction coefficient of the $\mathrm{i^{th}}$ compound ($\nu_i$ is positive for products and negative for reactants).
If none of the $c_{p,i}$ have a temperature dependence, we can rewrite the $1^{st}$ equation above as $$H_{T_f}=H_{T_i}+\Delta c_p\cdot (T_f-T_i)$$ If they do have a temperature dependence, you will have to know the functional form of $c_{p,i}$ with respect to temperature to integrate them.
So, in regards to your specific question, the bond enthalpy for an $\ce{H-H}$ bond will be different at different temperatures. However, it is worth noting that for highly energetic reactions like this, the contribution due to $c_{p,i}$ is usually fairly small in comparison.
