To solve the problem of finding the adjusted enthalpy for REACTION 3, we need to understand the process of reversing a chemical reaction and how it affects the enthalpy change, [tex]\(\Delta H\)[/tex].
Given:
[tex]\[
3:\, 2 C + 3 H_2 + \frac{1}{2} O_2 \rightarrow C_2H_5OH \, , \quad \Delta H = -278 \text{ kJ}
\][/tex]
To find the adjusted enthalpy when the reaction is reversed, we follow these steps:
1. Identify the original reaction and its enthalpy change:
The original reaction is:
[tex]\[
2 C + 3 H_2 + \frac{1}{2} O_2 \rightarrow C_2H_5OH
\][/tex]
with an enthalpy change, [tex]\(\Delta H\)[/tex], of [tex]\(-278 \text{kJ}\)[/tex].
2. Reverse the reaction:
Reversing the reaction means we swap the products and reactants:
[tex]\[
C_2H_5OH \rightarrow 2 C + 3 H_2 + \frac{1}{2} O_2
\][/tex]
3. Adjust the enthalpy change for the reversed reaction:
When a reaction is reversed, the sign of the enthalpy change is also reversed. Therefore, the [tex]\(\Delta H\)[/tex] for the reversed reaction will be the negative of the original enthalpy change.
The original [tex]\(\Delta H\)[/tex] is [tex]\(-278 \, \text{kJ}\)[/tex], so for the reversed reaction:
[tex]\[
\Delta H_{\text{reversed}} = -(-278 \, \text{kJ}) = +278 \, \text{kJ}
\][/tex]
Thus, the adjusted enthalpy for the reversed REACTION 3 is:
[tex]\[
\boxed{+278 \, \text{kJ}}
\][/tex]
