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Hess' Law \\
Combustion of Ethanol \\
Given the following thermochemical equations:
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Considering the given and goal reactions, how is \\
REACTION 1 manipulated to match the overall \\
reaction?
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[tex]$1: C + O_2 \rightarrow CO_2, \Delta H = -394 \text{ kJ}$[/tex] & It remains the same. \\
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[tex]$2: 3 H_2 + \frac{3}{2} O_2 \rightarrow 3 H_2O, \Delta H = -858 \text{ kJ}$[/tex] & It remains the same. \\
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[tex]$3: C_2H_5OH \rightarrow 2 C + 3 H_2 + \frac{1}{2} O_2, \Delta H = +278 \text{ kJ}$[/tex] & It is reversed. \\
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The goal is to use Hess's Law to calculate the standard \\
reaction enthalpy, [tex]$\Delta H$[/tex] for the reaction below.
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[tex]$C_2H_5OH + 3 O_2 \rightarrow 2 CO_2 + 3 H_2O$[/tex] & \\
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Sagot :

To determine the standard reaction enthalpy, ΔH, for the combustion of ethanol (C[tex]\(_2\)[/tex]H[tex]\(_5\)[/tex]OH), we can use Hess's Law. This law allows us to add up the enthalpy changes of multiple thermochemical equations to find the enthalpy change for an overall reaction. Let's examine the given equations and their manipulations:

### Given Thermochemical Equations and Their Manipulations
1. Equation 1: [tex]\(C + O_2 \rightarrow CO_2, \Delta H = -394 \text{ kJ}\)[/tex]
2. Equation 2: [tex]\(3 H_2 + \dfrac{3}{2} O_2 \rightarrow 3 H_2O, \Delta H = -858 \text{ kJ}\)[/tex]
3. Equation 3 (reversed): [tex]\(C_2H_5OH \rightarrow 2C + 3H_2 + \dfrac{1}{2} O_2, \Delta H = +278 \text{ kJ}\)[/tex]

The reversed form of Reaction 3:
[tex]\[2C + 3H_2 + \dfrac{1}{2} O_2 \rightarrow C_2H_5OH, \Delta H = -278 \text{ kJ}\][/tex]

### Target Reaction:
[tex]\[C_2H_5OH + 3 O_2 \rightarrow 2 CO_2 + 3 H_2O\][/tex]

### Step-by-Step Calculation:

1. Reaction 1: [tex]\(2C + 2O_2 \rightarrow 2CO_2\)[/tex]

Since the target reaction has [tex]\(2CO_2\)[/tex], we need to multiply Reaction 1 by 2:
[tex]\[ 2(C + O_2 \rightarrow CO_2), \quad \Delta H = 2 \times -394 \text{ kJ} = -788 \text{ kJ} \][/tex]

2. Reaction 2: [tex]\(3 H_2 + \dfrac{3}{2} O_2 \rightarrow 3 H_2O\)[/tex]

This matches the target reaction in terms of the [tex]\(3 H_2O\)[/tex] formed:
[tex]\[ 3 H_2 + \dfrac{3}{2} O_2 \rightarrow 3 H_2O, \quad \Delta H = -858 \text{ kJ} \][/tex]

3. Reaction 3 (reversed): [tex]\(C_2H_5OH + \dfrac{1}{2} O_2 \rightarrow 2C + 3H_2\)[/tex]

The reversed Reaction 3 is already properly aligned with the target reaction:
[tex]\[ 2C + 3H_2 + \dfrac{1}{2} O_2 \rightarrow C_2H_5OH, \quad \Delta H = -278 \text{ kJ} \][/tex]

### Combine the Reactions:
By summing all manipulated reactions, we get:
[tex]\[ \begin{aligned} & \left( 2C + 2O_2 \rightarrow 2CO_2, \Delta H_1 = -788 \text{ kJ} \right) \\ & + \left( 3 H_2 + \dfrac{3}{2} O_2 \rightarrow 3 H_2O, \Delta H_2 = -858 \text{ kJ} \right) \\ & + \left( C_2H_5OH + \dfrac{1}{2} O_2 \rightarrow 2C + 3 H_2, \Delta H_3 = -278 \text{ kJ} \right) \end{aligned} \][/tex]

### Enthalpy Change of the Target Reaction:
Summing the enthalpy changes of the individual steps:

[tex]\[ \Delta H_{\text{combustion, C_2H_5OH}} = -788 \text{ kJ} + -858 \text{ kJ} + -278 \text{ kJ} = -1924 \text{ kJ} \][/tex]

So the standard reaction enthalpy, ΔH, for the combustion of ethanol is:

[tex]\[ \Delta H_{\text{combustion, C_2H_5OH}} = -1530 \text{ kJ} \][/tex]

Therefore, [tex]\(\Delta H\)[/tex] for the reaction [tex]\(C_2H_5OH + 3 O_2 \rightarrow 2 CO_2 + 3 H_2O\)[/tex] is [tex]\(-1530\)[/tex] kJ.