Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
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.
[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.
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.