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Hess' Law - Combustion of Ethanol

Given the following thermochemical equations:
[tex]\[
\begin{array}{l}
1: C + O_2 \rightarrow CO_2, \Delta H = -394 \, \text{kJ} \\
2: 3 H_2 + \frac{3}{2} O_2 \rightarrow 3 H_2O, \Delta H = -858 \, \text{kJ} \\
3: C_2H_5OH \rightarrow 2 C + 3 H_2 + \frac{1}{2} O_2, \Delta H = +278 \, \text{kJ}
\end{array}
\][/tex]

The goal is to use Hess's Law to calculate the standard reaction enthalpy, [tex]\(\Delta H^{\circ}\)[/tex], for the reaction below:
[tex]\[
C_2H_5OH + 3 O_2 \rightarrow 2 CO_2 + 3 H_2O
\][/tex]

Considering the given and goal reactions, how is REACTION 1 manipulated to match the overall reaction?

A. It is reversed.
B. It remains the same.
C. It is tripled.
D. It is doubled.

Sagot :

GinnyAnswer
To determine how Reaction 1 is manipulated to match the overall reaction using Hess's Law, let's go through each given reaction and see how they contribute to forming the desired reaction:

1. [tex]\( C + O_2 \rightarrow CO_2, \Delta H = -394 \, \text{kJ} \)[/tex]
2. [tex]\( 3 H_2 + \frac{3}{2} O_2 \rightarrow 3 H_2O, \Delta H = -858 \, \text{kJ} \)[/tex]
3. [tex]\( C_2H_5OH \rightarrow 2 C + 3 H_2 + \frac{1}{2} O_2, \Delta H = +278 \, \text{kJ} \)[/tex]

The goal is to find the enthalpy change ([tex]\(\Delta H^{\circ}\)[/tex]) for the reaction:
[tex]\[ C_2H_5OH + 3 O_2 \rightarrow 2 CO_2 + 3 H_2O \][/tex]

First, consider the desired products and reactants in the goal reaction:
- We need 2 moles of [tex]\(CO_2\)[/tex].
- We need 3 moles of [tex]\(H_2O\)[/tex].
- We start with [tex]\(C_2H_5OH\)[/tex] and [tex]\(O_2\)[/tex].

To achieve this, let's rearrange the given reactions:

Step-by-step breakdown:

1. Start with Reaction 3, which involves [tex]\(C_2H_5OH\)[/tex]:
[tex]\[ C_2H_5OH \rightarrow 2 C + 3 H_2 + \frac{1}{2} O_2, \Delta H = +278 \, \text{kJ} \][/tex]
- This reaction must be reversed to match the [tex]\(C_2H_5OH\)[/tex] on the reactant side:
[tex]\[ 2 C + 3 H_2 + \frac{1}{2} O_2 \rightarrow C_2H_5OH, \Delta H = -278 \, \text{kJ} \][/tex]

2. Use Reaction 1 to form [tex]\(CO_2\)[/tex]:
[tex]\[ C + O_2 \rightarrow CO_2, \Delta H = -394 \, \text{kJ} \][/tex]
- We need 2 moles of [tex]\(CO_2\)[/tex], so Reaction 1 is used twice:
[tex]\[ 2(C + O_2 \rightarrow CO_2) \Rightarrow 2 C + 2 O_2 \rightarrow 2 CO_2, \Delta H = 2 \times -394 = -788 \, \text{kJ} \][/tex]

3. Use Reaction 2 to form [tex]\(H_2O\)[/tex]:
[tex]\[ 3 H_2 + \frac{3}{2} O_2 \rightarrow 3 H_2O, \Delta H = -858 \, \text{kJ} \][/tex]

Combine all the manipulated reactions:

1. [tex]\( 2 C + 3 H_2 + \frac{1}{2} O_2 \rightarrow C_2H_5OH, \Delta H = -278 \, \text{kJ} \)[/tex] (reversed Reaction 3)
2. [tex]\( 2 C + 2 O_2 \rightarrow 2 CO_2, \Delta H = -788 \, \text{kJ} \)[/tex] (twice Reaction 1)
3. [tex]\( 3 H_2 + \frac{3}{2} O_2 \rightarrow 3 H_2O, \Delta H = -858 \, \text{kJ} \)[/tex]

Note that these combinations produce:
[tex]\[ C_2H_5OH + 3 O_2 \rightarrow 2 CO_2 + 3 H_2O \][/tex]

Thus, Reaction 1 is used as is without any change to match the overall reaction. Therefore, the correct manipulation of Reaction 1 is:

- It remains the same.
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