To determine how Reaction 3 should be manipulated to match the overall reaction using Hess's Law, let's consider the goal reaction and the given thermochemical equations:
### Goal Reaction:
[tex]\[ C_2H_5OH + 3 O_2 \rightarrow 2 CO_2 + 3 H_2O \][/tex]
### Given Reactions:
1. [tex]\( \text{C} + \text{O}_2 \rightarrow \text{CO}_2, \Delta H = -394 \text{kJ} \)[/tex]
2. [tex]\( \text{H}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{H}_2\text{O}, \Delta H = -286 \text{kJ} \)[/tex]
3. [tex]\( 2 \text{C} + 3 \text{H}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{C}_2\text{H}_5\text{OH}, \Delta H = -278 \text{kJ} \)[/tex]
### Strategy:
To use these reactions to achieve the goal reaction, we need to manipulate Reaction 3 to get ethanol (C[tex]\(_2\)[/tex]H[tex]\(_5\)[/tex]OH) on the reactant side.
1. Initial Form of Reaction 3:
[tex]\( 2 \text{C} + 3 \text{H}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{C}_2\text{H}_5\text{OH}, \Delta H = -278 \text{kJ} \)[/tex]
2. Required Manipulation:
Since the goal reaction has ethanol on the reactant side, we need to reverse Reaction 3. Reversing the reaction will also reverse the sign of the enthalpy change.
3. Reversed Reaction 3:
[tex]\( \text{C}_2\text{H}_5\text{OH} \rightarrow 2 \text{C} + 3 \text{H}_2 + \frac{1}{2} \text{O}_2, \Delta H = +278 \text{kJ} \)[/tex]
4. Balancing with Goal Reaction:
The reversed Reaction 3 now matches the ethanol component of the goal reaction. We don't need to scale it, as the coefficients already fit the goal reaction.
In conclusion, Reaction 3 should be reversed to match the overall reaction:
[tex]\[ \boxed{\text{It is reversed.}} \][/tex]
Thus, the correct manipulation of Reaction 3 is to reverse it.
