To solve the given problem using Hess's Law, let’s follow these steps systematically:
### Given Reactions:
1. [tex]\( C + O_2 \rightarrow CO_2 \)[/tex], [tex]\( \Delta H = -394 \)[/tex] kJ \\
2. [tex]\( H_2 + \frac{1}{2} O_2 \rightarrow H_2O \)[/tex], [tex]\( \Delta H = -286 \)[/tex] kJ \\
3. [tex]\( C_2H_5OH \rightarrow 2C + 3H_2 + \frac{1}{2} O_2 \)[/tex], [tex]\( \Delta H = +278 \)[/tex] kJ
### Goal Reaction:
[tex]\( C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O \)[/tex]
### Steps to achieve the goal reaction:
1. Reverse Reaction 3:
[tex]\[
2C + 3H_2 + \frac{1}{2} O_2 \rightarrow C_2H_5OH \quad \Delta H = -278 \text{ kJ}
\][/tex]
2. Manipulate the individual reactions to match the goal reaction:
- Use Reaction 1 as is:
[tex]\[
C + O_2 \rightarrow CO_2 \quad \Delta H = -394 \text{ kJ}
\][/tex]
- Use Reaction 2 as is:
[tex]\[
H_2 + \frac{1}{2} O_2 \rightarrow H_2O \quad \Delta H = -286 \text{ kJ}
\][/tex]
3. Combine the reactions:
- We need 2 moles of Reaction 1:
[tex]\[
2(C + O_2 \rightarrow CO_2) \quad \Delta H = 2 \times -394 = -788 \text{ kJ}
\][/tex]
- We need 3 moles of Reaction 2:
[tex]\[
3(H_2 + \frac{1}{2} O_2 \rightarrow H_2O) \quad \Delta H = 3 \times -286 = -858 \text{ kJ}
\][/tex]
- We also include the reversed Reaction 3:
[tex]\[
2C + 3H_2 + \frac{1}{2} O_2 \rightarrow C_2H_5OH \quad \Delta H = -278 \text{ kJ}
\][/tex]
### Combine and sum the enthalpies:
Combining these modified equations:
1. [tex]\( 2C + 2O_2 \rightarrow 2CO_2 \)[/tex] \quad ([tex]\( \Delta H = -788 \text{ kJ} \)[/tex])
2. [tex]\( 3H_2 + \frac{3}{2} O_2 \rightarrow 3H_2O \)[/tex] \quad ([tex]\( \Delta H = -858 \text{ kJ} \)[/tex])
3. [tex]\( 2C + 3H_2 + \frac{1}{2} O_2 \rightarrow C_2H_5OH \)[/tex] \quad ([tex]\( \Delta H = -278 \text{ kJ} \)[/tex])
By summing the reactions, the undesired products (2C, 3H_2, [tex]\(\frac{1}{2} O_2\)[/tex]) cancel out:
[tex]\[
-788 + -858 + -278 = -1924 \quad \text{(but this is a common mistake in adding, correct should be -1368 is sum of -788, -858 and +278.)}
\][/tex]
Therefore, the standard enthalpy change for the reaction is:
[tex]\[
\Delta H^\circ = -1368 \text{ kJ}
\][/tex]
### How is Reaction 2 manipulated to match the overall reaction?
- Reaction 2 remains the same.
- It is not reversed, doubled, or tripled.
The final answer is:
It remains the same.
