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Hess' Law: Combustion of Ethanol

Given the following thermochemical equations:
[tex]\[
\begin{array}{l}
\text{1: } C + O_2 \rightarrow CO_2, \Delta H = -394 \, \text{kJ} \\
\text{2: } H_2 + \frac{1}{2} O_2 \rightarrow H_2O, \Delta H = -286 \, \text{kJ} \\
\text{3: } C_2H_5OH \rightarrow 2C + 3H_2 + \frac{1}{2} O_2, \Delta H = +278 \, \text{kJ}
\end{array}
\][/tex]

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

Considering the given and goal reactions, how is Reaction 2 manipulated to match the overall reaction?

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

Sagot :

GinnyAnsweridn
To solve the given problem using Hess's Law, we need to use the provided thermochemical equations to find the standard reaction enthalpy, [tex]\(\Delta H\)[/tex], for the target reaction:

[tex]\[ C_2H_5OH + 3 O_2 \rightarrow 2 CO_2 + 3 H_2O \][/tex]

Given thermochemical equations:
1. [tex]\( C + O_2 \rightarrow CO_2, \Delta H = -394 \, \text{kJ} \)[/tex]
2. [tex]\( H_2 + \frac{1}{2} O_2 \rightarrow H_2O, \Delta H = -286 \, \text{kJ} \)[/tex]
3. [tex]\( C_2H_5OH \rightarrow 2 C + 3 H_2 + \frac{1}{2} O_2, \Delta H = +278 \, \text{kJ} \)[/tex]

We manipulate these equations to match the target reaction:

### Step-by-Step Solution:

1. Reverse Equation 3:
[tex]\[ C_2H_5OH \rightarrow 2 C + 3 H_2 + \frac{1}{2} O_2, \Delta H = +278 \, \text{kJ} \][/tex]
When reversed:
[tex]\[ 2 C + 3 H_2 + \frac{1}{2} O_2 \rightarrow C_2H_5OH, \Delta H = -278 \, \text{kJ} \][/tex]

2. Double Equation 1:
[tex]\[ C + O_2 \rightarrow CO_2, \Delta H = -394 \, \text{kJ} \][/tex]
When doubled:
[tex]\[ 2 C + 2 O_2 \rightarrow 2 CO_2, \Delta H = 2 \times -394 \, \text{kJ} = -788 \, \text{kJ} \][/tex]

3. Triple Equation 2:
[tex]\[ H_2 + \frac{1}{2} O_2 \rightarrow H_2O, \Delta H = -286 \, \text{kJ} \][/tex]
When tripled:
[tex]\[ 3 H_2 + \frac{3}{2} O_2 \rightarrow 3 H_2O, \Delta H = 3 \times -286 \, \text{kJ} = -858 \, \text{kJ} \][/tex]

4. Combine the manipulated equations:
Reversed Equation 3:
[tex]\[ 2 C + 3 H_2 + \frac{1}{2} O_2 \rightarrow C_2H_5OH, \Delta H = -278 \, \text{kJ} \][/tex]

Doubled Equation 1:
[tex]\[ 2 C + 2 O_2 \rightarrow 2 CO_2, \Delta H = -788 \, \text{kJ} \][/tex]

Tripled Equation 2:
[tex]\[ 3 H_2 + \frac{3}{2} O_2 \rightarrow 3 H_2O, \Delta H = -858 \, \text{kJ} \][/tex]

Adding these reactions gives the overall reaction:
[tex]\[ C_2H_5OH + 3 O_2 \rightarrow 2 CO_2 + 3 H_2O \][/tex]

The enthalpy changes sum up as follows:
[tex]\[ \Delta H_{total} = \Delta H_3^{reversed} + \Delta H_1^{doubled} + \Delta H_2^{tripled} \][/tex]
[tex]\[ \Delta H_{total} = -278 \, \text{kJ} + (-788 \, \text{kJ}) + (-858 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{total} = -1924 \, \text{kJ} \][/tex]

### Additional Question: How is Reaction 2 manipulated?
Examining the manipulation, Reaction 2 is used as:
[tex]\[ H_2 + \frac{1}{2} O_2 \rightarrow H_2O, \Delta H = -286 \, \text{kJ} \][/tex]
We triple it (to get 3 times the enthalpy change and reactants/products).

Therefore, Reaction 2 is tripled.
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