To find the enthalpy change for the reversed reaction, we need to understand a basic principle in thermochemistry. When a chemical reaction is reversed, the sign of the enthalpy change ([tex]\(\Delta H\)[/tex]) is also reversed.
Given the original reaction:
[tex]\[2 \text{H}_2 (g) + \text{O}_2 (g) \rightarrow 2 \text{H}_2 \text{O} (g) \quad \Delta H^{\circ} = -484.0 \, \frac{kJ}{mol}\][/tex]
We are interested in the enthalpy change for the reversed reaction:
[tex]\[2 \text{H}_2 \text{O} (g) \rightarrow 2 \text{H}_2 (g) + \text{O}_2 (g)\][/tex]
For the reversed reaction, the enthalpy change will have the opposite sign of the original reaction.
Since the original [tex]\(\Delta H^{\circ}\)[/tex] is [tex]\(-484.0 \, \frac{kJ}{mol}\)[/tex], the [tex]\(\Delta H^{\circ}\)[/tex] for the reversed reaction will be:
[tex]\[ \Delta H^{\circ} = +484.0 \, \frac{kJ}{mol} \][/tex]
Thus, the enthalpy for the reversed reaction is:
[tex]\[ +484.0 \, \frac{kJ}{mol} \][/tex]
This answer accounts for significant figures, given that the original enthalpy was provided with four significant figures.
