Certainly! Let's start by understanding that enthalpy change ([tex]\(\Delta H\)[/tex]) for a reaction is a measure of the heat released or absorbed during the process. In the given problem, we have the following chemical reaction:
[tex]\[ 2 H_2(g) + O_2(g) \rightarrow 2 H_2O(g) \][/tex]
with an enthalpy change:
[tex]\[ \Delta H^{\circ} = -484.0 \frac{ kJ }{ mol } \][/tex]
This indicates that the reaction releases 484.0 kJ of energy per mole of reaction, making it an exothermic reaction (since the sign of [tex]\(\Delta H\)[/tex] is negative).
However, the problem states that we need the enthalpy change for the reversed reaction. Reversing a chemical reaction involves reversing the sign of the enthalpy change because the direction of heat flow is reversed.
In the forward reaction:
[tex]\[ 2 H_2(g) + O_2(g) \rightarrow 2 H_2O(g) \][/tex]
[tex]\[ \Delta H^{\circ} = -484.0 \frac{ kJ }{ mol } \][/tex]
In the reverse reaction:
[tex]\[ 2 H_2O(g) \rightarrow 2 H_2(g) + O_2(g) \][/tex]
the enthalpy change becomes:
[tex]\[ \Delta H^{\circ} = +484.0 \frac{ kJ }{ mol } \][/tex]
This means that 484.0 kJ per mole of energy is absorbed for the reversed process, making it an endothermic reaction (since the sign of [tex]\(\Delta H\)[/tex] is positive).
Therefore, the enthalpy for the modified reaction is:
[tex]\[ +484.0 \, \frac{kJ}{mol} \][/tex]
Hence the final answer is [tex]\( +484.0 \, kJ/mol \)[/tex], considering the appropriate significant figures.
