IDNLearn.com provides a collaborative environment for finding accurate answers. Our community provides accurate and timely answers to help you understand and solve any issue.
Sagot :
To determine the enthalpy diagram for this system using Hess's law, we'll analyze the given reactions one by one and ensure we segment the problem into clear steps.
### Step-by-Step Solution
1. Consider the Given Intermediate Reactions:
[tex]\[ \begin{array}{ll} \text{Reaction 1:} & CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g) \quad \Delta H_1 = -802 \, \text{kJ} \\ \text{Reaction 2:} & 2H_2O(g) \rightarrow 2H_2O(l) \quad \Delta H_2 = -88 \, \text{kJ} \end{array} \][/tex]
2. Identify the Overall Reaction:
[tex]\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad \Delta H_{\text{overall}} = -890 \, \text{kJ} \][/tex]
3. Apply Hess's Law:
According to Hess's law, the change in enthalpy for the overall reaction is the sum of enthalpy changes for the steps into which the reaction can be divided.
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + \Delta H_2 \][/tex]
4. Determine the Enthalpy Change Step-by-Step:
From the intermediate reactions, adding the enthalpy changes should equate to the enthalpy change of the overall reaction.
[tex]\[ -802 \, \text{kJ} + (-88 \, \text{kJ}) = -890 \, \text{kJ} \][/tex]
This validates that our intermediate reactions and the overall reaction align correctly.
5. Construct the Enthalpy Diagram:
The enthalpy diagram will show the progression of enthalpy changes from reactants to intermediates to products:
- Initial state (Reactants): [tex]\( CH_4(g) + 2O_2(g) \)[/tex]
- First intermediate step: Formation of [tex]\( CO_2(g) + 2H_2O(g) \)[/tex]
- Enthalpy change: [tex]\( \Delta H_1 = -802 \, \text{kJ} \)[/tex]
- Second intermediate step: Condensation of [tex]\( 2H_2O(g) \)[/tex] to [tex]\( 2H_2O(l) \)[/tex]
- Enthalpy change: [tex]\( \Delta H_2 = -88 \, \text{kJ} \)[/tex]
- Final state (Products): [tex]\( CO_2(g) + 2H_2O(l) \)[/tex]
- Total enthalpy change: [tex]\( \Delta H_{\text{overall}} = -890 \, \text{kJ} \)[/tex]
### Enthalpy Diagram Representation
The enthalpy diagram can be drawn as follows:
1. Start at some initial enthalpy level for the reactants [tex]\( CH_4(g) + 2O_2(g) \)[/tex].
2. Draw a downward arrow to a lower enthalpy level representing the products of the first reaction [tex]\( CO_2(g) + 2H_2O(g) \)[/tex] with an enthalpy change of [tex]\( -802 \, \text{kJ} \)[/tex].
3. From this intermediate level, draw another downward arrow to an even lower enthalpy level representing the final products [tex]\( CO_2(g) + 2H_2O(l) \)[/tex], adding the additional enthalpy change of [tex]\( -88 \, \text{kJ} \)[/tex].
4. The overall drop in enthalpy from the initial reactants to the final products should be [tex]\( -890 \, \text{kJ} \)[/tex].
This diagram visually shows how the enthalpy of the system changes step-by-step. Here is a rough sketch of the diagram:
[tex]\[ \begin{array}{c} \text{Reactants: } CH_4(g) + 2O_2(g) \\ |\Delta H_1 = -802 \, \text{kJ}| \\ \downarrow \\ \text{Intermediate: } CO_2(g) + 2H_2O(g) \\ |\Delta H_2 = -88 \, \text{kJ}| \\ \downarrow \\ \text{Products: } CO_2(g) + 2H_2O(l) \end{array} \][/tex]
### Summary
Applying Hess's law, we confirmed the consistency of the given enthalpy changes with the overall reaction. The enthalpy diagram depicted above accurately represents these changes in sequential steps.
### Step-by-Step Solution
1. Consider the Given Intermediate Reactions:
[tex]\[ \begin{array}{ll} \text{Reaction 1:} & CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g) \quad \Delta H_1 = -802 \, \text{kJ} \\ \text{Reaction 2:} & 2H_2O(g) \rightarrow 2H_2O(l) \quad \Delta H_2 = -88 \, \text{kJ} \end{array} \][/tex]
2. Identify the Overall Reaction:
[tex]\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad \Delta H_{\text{overall}} = -890 \, \text{kJ} \][/tex]
3. Apply Hess's Law:
According to Hess's law, the change in enthalpy for the overall reaction is the sum of enthalpy changes for the steps into which the reaction can be divided.
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + \Delta H_2 \][/tex]
4. Determine the Enthalpy Change Step-by-Step:
From the intermediate reactions, adding the enthalpy changes should equate to the enthalpy change of the overall reaction.
[tex]\[ -802 \, \text{kJ} + (-88 \, \text{kJ}) = -890 \, \text{kJ} \][/tex]
This validates that our intermediate reactions and the overall reaction align correctly.
5. Construct the Enthalpy Diagram:
The enthalpy diagram will show the progression of enthalpy changes from reactants to intermediates to products:
- Initial state (Reactants): [tex]\( CH_4(g) + 2O_2(g) \)[/tex]
- First intermediate step: Formation of [tex]\( CO_2(g) + 2H_2O(g) \)[/tex]
- Enthalpy change: [tex]\( \Delta H_1 = -802 \, \text{kJ} \)[/tex]
- Second intermediate step: Condensation of [tex]\( 2H_2O(g) \)[/tex] to [tex]\( 2H_2O(l) \)[/tex]
- Enthalpy change: [tex]\( \Delta H_2 = -88 \, \text{kJ} \)[/tex]
- Final state (Products): [tex]\( CO_2(g) + 2H_2O(l) \)[/tex]
- Total enthalpy change: [tex]\( \Delta H_{\text{overall}} = -890 \, \text{kJ} \)[/tex]
### Enthalpy Diagram Representation
The enthalpy diagram can be drawn as follows:
1. Start at some initial enthalpy level for the reactants [tex]\( CH_4(g) + 2O_2(g) \)[/tex].
2. Draw a downward arrow to a lower enthalpy level representing the products of the first reaction [tex]\( CO_2(g) + 2H_2O(g) \)[/tex] with an enthalpy change of [tex]\( -802 \, \text{kJ} \)[/tex].
3. From this intermediate level, draw another downward arrow to an even lower enthalpy level representing the final products [tex]\( CO_2(g) + 2H_2O(l) \)[/tex], adding the additional enthalpy change of [tex]\( -88 \, \text{kJ} \)[/tex].
4. The overall drop in enthalpy from the initial reactants to the final products should be [tex]\( -890 \, \text{kJ} \)[/tex].
This diagram visually shows how the enthalpy of the system changes step-by-step. Here is a rough sketch of the diagram:
[tex]\[ \begin{array}{c} \text{Reactants: } CH_4(g) + 2O_2(g) \\ |\Delta H_1 = -802 \, \text{kJ}| \\ \downarrow \\ \text{Intermediate: } CO_2(g) + 2H_2O(g) \\ |\Delta H_2 = -88 \, \text{kJ}| \\ \downarrow \\ \text{Products: } CO_2(g) + 2H_2O(l) \end{array} \][/tex]
### Summary
Applying Hess's law, we confirmed the consistency of the given enthalpy changes with the overall reaction. The enthalpy diagram depicted above accurately represents these changes in sequential steps.
Your engagement is important to us. Keep sharing your knowledge and experiences. Let's create a learning environment that is both enjoyable and beneficial. Your search for solutions ends here at IDNLearn.com. Thank you for visiting, and come back soon for more helpful information.
