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Consider the following intermediate reactions:

[tex]\[
\begin{array}{ll}
CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g) & \Delta H_1 = -802 \, \text{kJ} \\
2 H_2O(g) \rightarrow 2 H_2O(l) & \Delta H_2 = -88 \, \text{kJ}
\end{array}
\][/tex]

The overall chemical reaction is as follows:

[tex]\[
CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \quad \Delta H = -890 \, \text{kJ}
\][/tex]

What is the correct enthalpy diagram using Hess's Law for this system?

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.