To determine the adjusted enthalpy for Reaction 3 after it has been reversed, we need to follow the principles of Hess' Law. Hess' Law states that the enthalpy change of a whole reaction is the sum of the enthalpy changes of its individual steps. One important rule when applying Hess' Law is that when the direction of a reaction is reversed, the sign of the enthalpy change (ΔH) is also reversed.
Given Reaction 3:
[tex]\[ 2C + 3H_2 + \frac{1}{2} O_2 \rightarrow C_2H_5OH, \quad \Delta H = -278 \text{ kJ} \][/tex]
This reaction has an enthalpy change of [tex]\(-278 \text{ kJ}\)[/tex]. To reverse the reaction means to move from the products to the reactants:
[tex]\[ C_2H_5OH \rightarrow 2C + 3H_2 + \frac{1}{2} O_2 \][/tex]
When reversing the reaction, the enthalpy change will also reverse its sign. Therefore, if the enthalpy change for the forward reaction (from reactants to products) is [tex]\(-278 \text{ kJ}\)[/tex], the enthalpy change for the reverse reaction (from products to reactants) will be:
[tex]\[ +278 \text{ kJ} \][/tex]
Thus, the adjusted enthalpy change for Reaction 3 after reversing it is:
[tex]\[ \boxed{+278 \text{ kJ}} \][/tex]
So, the correct answer is:
[tex]\[ +278 \text{ kJ} \][/tex]
