Let's walk through the provided problem step by step:
### Part 1: Energy released during the formation of [tex]\(1 \text{ mol of } \mathrm{H_2O(g)}\)[/tex]
The balanced chemical equation is given by:
[tex]\[
2 \,\mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow 2 \,\mathrm{H_2O(g)}
\][/tex]
with an enthalpy change [tex]\( \Delta H_{\text{rxn}} = -483.64 \, \text{kJ} \)[/tex]. This enthalpy change indicates that 483.64 kJ of energy is released during the formation of 2 moles of [tex]\( \mathrm{H_2O(g)} \)[/tex].
To find out how much energy is released during the formation of 1 mole of [tex]\( \mathrm{H_2O(g)} \)[/tex], we simply divide the total energy released by the number of moles of [tex]\( \mathrm{H_2O(g)} \)[/tex]:
[tex]\[
\frac{-483.64 \, \text{kJ}}{2} = -241.82 \, \text{kJ}
\][/tex]
So, during the formation of 1 mole of [tex]\( \mathrm{H_2O(g)} \)[/tex], [tex]\( -241.82 \, \text{kJ} \)[/tex] of energy is released.
### Part 2: Energy required for the reverse reaction
The reverse reaction is:
[tex]\[
2 \, \mathrm{H_2O(g)} \rightarrow 2 \, \mathrm{H_2(g)} + \mathrm{O_2(g)}
\][/tex]
The energy associated with the reverse reaction is the same as the energy of the forward reaction but with the opposite sign. Thus, the enthalpy change for the reverse reaction is positive. This is because energy needs to be added to break the bonds in [tex]\( \mathrm{H_2O(g)} \)[/tex] to form [tex]\( \mathrm{H_2(g)} \)[/tex] and [tex]\( \mathrm{O_2(g)} \)[/tex].
Therefore, the amount of energy that must be added for the reverse reaction is:
[tex]\[
+483.64 \, \text{kJ}
\][/tex]
So, the energy that must be added to the reaction [tex]\( 2 \,\mathrm{H_2O(g)} \rightarrow 2 \,\mathrm{H_2(g)} + \mathrm{O_2(g)} \)[/tex] is [tex]\( 483.64 \, \text{kJ} \)[/tex].
