To determine which reaction needs to be reversed to achieve the overall reaction, we need to follow these steps carefully:
1. Identify the given reactions and their enthalpy changes:
- Reaction 1: [tex]\( C_2H_5OH + 3 O_2 \rightarrow 2 CO_2 + 3 H_2O \)[/tex], [tex]\( \Delta H = -1370 \, \text{kJ/mol} \)[/tex]
- Reaction 2: [tex]\( C_2H_4 + 3 O_2 \rightarrow 2 CO_2 + 2 H_2O \)[/tex], [tex]\( \Delta H = -1410 \, \text{kJ/mol} \)[/tex]
2. Write the overall reaction:
- Overall reaction: [tex]\( C_2H_4 + H_2O \rightarrow C_2H_5OH \)[/tex]
3. Adjust reactions to form the overall reaction:
- To form the overall reaction from the given reactions, we need to reverse one of them so that the products of one reaction can cancel with the reactants of the other.
4. Reverse one of the reactions:
- We will reverse Reaction 1 and examine the combinations.
- Reversed Reaction 1: [tex]\( 2 CO_2 + 3 H_2O \rightarrow C_2H_5OH + 3 O_2 \)[/tex], [tex]\( \Delta H = +1370 \, \text{kJ/mol} \)[/tex]
5. Add the reversed Reaction 1 to Reaction 2:
- Reaction 2: [tex]\( C_2H_4 + 3 O_2 \rightarrow 2 CO_2 + 2 H_2O \)[/tex], [tex]\( \Delta H = -1410 \, \text{kJ/mol} \)[/tex]
- Reversed Reaction 1: [tex]\( 2 CO_2 + 3 H_2O \rightarrow C_2H_5OH + 3 O_2 \)[/tex], [tex]\( \Delta H = +1370 \, \text{kJ/mol} \)[/tex]
Adding these together:
[tex]\[
\begin{aligned}
(C_2H_4 + 3 O_2) + (2 CO_2 + 3 H_2O) &\rightarrow (2 CO_2 + 2 H_2O) + (C_2H_5OH + 3 O_2) \\
C_2H_4 + 3 O_2 + 2 CO_2 + 3 H_2O &\rightarrow 2 CO_2 + 2 H_2O + C_2H_5OH + 3 O_2
\end{aligned}
\][/tex]
6. Cancel the common species on both sides:
- Cancel [tex]\( 3 O_2 \)[/tex], [tex]\( 2 CO_2 \)[/tex], and [tex]\( 2 H_2O \)[/tex]:
[tex]\[
C_2H_4 + H_2O \rightarrow C_2H_5OH
\][/tex]
This matches the overall reaction we need: [tex]\( C_2H_4 + H_2O \rightarrow C_2H_5OH \)[/tex].
Therefore, to achieve the overall reaction from the given reactions, we need to reverse Reaction 1.
