Answer :
To find the standard reaction enthalpy (ΔH) for the combustion of ethanol using Hess's Law, we start by manipulating the given thermochemical equations to match the overall goal reaction. Here’s a detailed step-by-step solution:
1. Given Reactions:
[tex]\[ \begin{align*} 1. & \quad C + O_2 \rightarrow CO_2, \quad \Delta H_1 = -394 \text{kJ} \\ 2. & \quad H_2 + \frac{1}{2}O_2 \rightarrow H_2O, \quad \Delta H_2 = -286 \text{kJ} \\ 3. & \quad 2C + 3H_2 + \frac{1}{2}O_2 \rightarrow C_2H_5OH, \quad \Delta H_3 = -278 \text{kJ} \end{align*} \][/tex]
2. Goal Reaction:
[tex]\[ C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O \][/tex]
3. Manipulating Reaction 3:
To get [tex]\(C_2H_5OH\)[/tex] on the reactant side, we need to reverse Reaction 3:
[tex]\[ C_2H_5OH \rightarrow 2C + 3H_2 + \frac{1}{2}O_2 \][/tex]
When we reverse a reaction, we also reverse the sign of ΔH:
[tex]\[ \Delta H_{3,\text{reversed}} = +278 \text{kJ} \][/tex]
4. Using Hess’s Law:
To achieve the goal reaction, we add up the manipulated reactions. Let’s combine the following:
- Reaction 1: [tex]\(C + O_2 \rightarrow CO_2, \quad \Delta H_1 = -394 \text{kJ} \)[/tex]
- Reaction 2 multiplied by 3 (to get 3 [tex]\(H_2O\)[/tex]):
[tex]\[ 3 \left(H_2 + \frac{1}{2} O_2 \rightarrow H_2O\right) \quad \text{which results in} \quad 3H_2 + \frac{3}{2}O_2 \rightarrow 3H_2O \][/tex]
The enthalpy change for this multiplied reaction:
[tex]\[ \Delta H_2 \times 3 = -286 \times 3 = -858 \text{kJ} \][/tex]
- Reversed Reaction 3:
[tex]\[ C_2H_5OH \rightarrow 2C + 3H_2 + \frac{1}{2}O_2, \quad \Delta H_{3,\text{reversed}} = +278 \text{kJ} \][/tex]
5. Summing Up:
Combine the enthalpy changes to get the standard enthalpy change for the goal reaction:
[tex]\[ \Delta H_{\text{goal reaction}} = \Delta H_1 + \Delta H_{2,\text{tripled}} + \Delta H_{3,\text{reversed}} \][/tex]
[tex]\[ \Delta H_{\text{goal reaction}} = -394 + (-858) + 278 = -974 \text{kJ} \][/tex]
The standard enthalpy change (ΔH) for the combustion of ethanol, [tex]\(C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O\)[/tex], is:
[tex]\[ \boxed{-974 \text{kJ}} \][/tex]
1. Given Reactions:
[tex]\[ \begin{align*} 1. & \quad C + O_2 \rightarrow CO_2, \quad \Delta H_1 = -394 \text{kJ} \\ 2. & \quad H_2 + \frac{1}{2}O_2 \rightarrow H_2O, \quad \Delta H_2 = -286 \text{kJ} \\ 3. & \quad 2C + 3H_2 + \frac{1}{2}O_2 \rightarrow C_2H_5OH, \quad \Delta H_3 = -278 \text{kJ} \end{align*} \][/tex]
2. Goal Reaction:
[tex]\[ C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O \][/tex]
3. Manipulating Reaction 3:
To get [tex]\(C_2H_5OH\)[/tex] on the reactant side, we need to reverse Reaction 3:
[tex]\[ C_2H_5OH \rightarrow 2C + 3H_2 + \frac{1}{2}O_2 \][/tex]
When we reverse a reaction, we also reverse the sign of ΔH:
[tex]\[ \Delta H_{3,\text{reversed}} = +278 \text{kJ} \][/tex]
4. Using Hess’s Law:
To achieve the goal reaction, we add up the manipulated reactions. Let’s combine the following:
- Reaction 1: [tex]\(C + O_2 \rightarrow CO_2, \quad \Delta H_1 = -394 \text{kJ} \)[/tex]
- Reaction 2 multiplied by 3 (to get 3 [tex]\(H_2O\)[/tex]):
[tex]\[ 3 \left(H_2 + \frac{1}{2} O_2 \rightarrow H_2O\right) \quad \text{which results in} \quad 3H_2 + \frac{3}{2}O_2 \rightarrow 3H_2O \][/tex]
The enthalpy change for this multiplied reaction:
[tex]\[ \Delta H_2 \times 3 = -286 \times 3 = -858 \text{kJ} \][/tex]
- Reversed Reaction 3:
[tex]\[ C_2H_5OH \rightarrow 2C + 3H_2 + \frac{1}{2}O_2, \quad \Delta H_{3,\text{reversed}} = +278 \text{kJ} \][/tex]
5. Summing Up:
Combine the enthalpy changes to get the standard enthalpy change for the goal reaction:
[tex]\[ \Delta H_{\text{goal reaction}} = \Delta H_1 + \Delta H_{2,\text{tripled}} + \Delta H_{3,\text{reversed}} \][/tex]
[tex]\[ \Delta H_{\text{goal reaction}} = -394 + (-858) + 278 = -974 \text{kJ} \][/tex]
The standard enthalpy change (ΔH) for the combustion of ethanol, [tex]\(C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O\)[/tex], is:
[tex]\[ \boxed{-974 \text{kJ}} \][/tex]
