Answer :
To create the correct enthalpy diagram for the system using Hess's Law, we need to use the given enthalpy changes (\(\Delta H\)) for the intermediate reactions and the overall reaction. Here’s a step-by-step guide to visualize the process:
### Step 1: Understand the reactions and their enthalpies
We are given two intermediate reactions and one overall reaction:
1. Intermediate Reaction 1:
[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g) \quad \Delta H_1 = -802 \, \text{kJ} \][/tex]
2. Intermediate Reaction 2:
[tex]\[ 2 H_2O(g) \rightarrow 2 H_2O(l) \quad \Delta H_2 = -88 \, \text{kJ} \][/tex]
3. Overall Reaction:
[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \quad \Delta H = -890 \, \text{kJ} \][/tex]
### Step 2: Construct the Enthalpy Diagram
To use Hess's Law, we should construct an enthalpy diagram that traces the energies involved in each reaction step:
1. Initial State: Reactants
Begin by noting the enthalpy of the reactants \(CH_4(g)\) and \(2 O_2(g)\).
2. First Intermediate State
The first reaction takes us from \(CH_4(g)\) and \(2 O_2(g)\) to \(CO_2(g) + 2 H_2O(g)\) with \(\Delta H_1 = -802 \, \text{kJ}\).
- Indicate this step with a downward arrow where the initial enthalpy decreases by 802 kJ.
3. Second Intermediate State
The second reaction takes us from \(2 H_2O(g)\) to \(2 H_2O(l)\) with \(\Delta H_2 = -88 \, \text{kJ}\).
- Add another downward arrow beginning from the first intermediate state which further decreases the enthalpy by 88 kJ.
4. Final State: Products
The products of the overall reaction are \(CO_2(g)\) and \(2 H_2O(l)\). The enthalpy change for the overall reaction is \(\Delta H = -890 \, \text{kJ}\).
- The final state should be reflected in reaching this total enthalpy change from the initial state.
### Step 3: Verify Consistency with Hess's Law
Hess's Law states that the total enthalpy change for the overall reaction should be the sum of the enthalpy changes for the intermediate reactions:
[tex]\[ \Delta H_\text{total} = \Delta H_1 + \Delta H_2 \][/tex]
Let's calculate:
[tex]\[ \Delta H_\text{total} = -802 \, \text{kJ} + (-88 \, \text{kJ}) = -890 \, \text{kJ} \][/tex]
### Conclusion of Diagram
Given our step-by-step tracing of the enthalpy changes:
- Start at the initial enthalpy level,
- Drop by 802 kJ for the first intermediate reaction,
- Drop further by 88 kJ for the conversion of \(2 H_2O(g) \rightarrow 2 H_2O(l)\),
- End at a total enthalpy drop of 890 kJ from the starting point.
The enthalpy diagram should reflect these steps, forming a consistent path from reactants to products, with appropriate intermediate steps.
This detailed step-by-step process ensures that the enthalpy diagram aligns with Hess's Law for this chemical system.
### Step 1: Understand the reactions and their enthalpies
We are given two intermediate reactions and one overall reaction:
1. Intermediate Reaction 1:
[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g) \quad \Delta H_1 = -802 \, \text{kJ} \][/tex]
2. Intermediate Reaction 2:
[tex]\[ 2 H_2O(g) \rightarrow 2 H_2O(l) \quad \Delta H_2 = -88 \, \text{kJ} \][/tex]
3. Overall Reaction:
[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \quad \Delta H = -890 \, \text{kJ} \][/tex]
### Step 2: Construct the Enthalpy Diagram
To use Hess's Law, we should construct an enthalpy diagram that traces the energies involved in each reaction step:
1. Initial State: Reactants
Begin by noting the enthalpy of the reactants \(CH_4(g)\) and \(2 O_2(g)\).
2. First Intermediate State
The first reaction takes us from \(CH_4(g)\) and \(2 O_2(g)\) to \(CO_2(g) + 2 H_2O(g)\) with \(\Delta H_1 = -802 \, \text{kJ}\).
- Indicate this step with a downward arrow where the initial enthalpy decreases by 802 kJ.
3. Second Intermediate State
The second reaction takes us from \(2 H_2O(g)\) to \(2 H_2O(l)\) with \(\Delta H_2 = -88 \, \text{kJ}\).
- Add another downward arrow beginning from the first intermediate state which further decreases the enthalpy by 88 kJ.
4. Final State: Products
The products of the overall reaction are \(CO_2(g)\) and \(2 H_2O(l)\). The enthalpy change for the overall reaction is \(\Delta H = -890 \, \text{kJ}\).
- The final state should be reflected in reaching this total enthalpy change from the initial state.
### Step 3: Verify Consistency with Hess's Law
Hess's Law states that the total enthalpy change for the overall reaction should be the sum of the enthalpy changes for the intermediate reactions:
[tex]\[ \Delta H_\text{total} = \Delta H_1 + \Delta H_2 \][/tex]
Let's calculate:
[tex]\[ \Delta H_\text{total} = -802 \, \text{kJ} + (-88 \, \text{kJ}) = -890 \, \text{kJ} \][/tex]
### Conclusion of Diagram
Given our step-by-step tracing of the enthalpy changes:
- Start at the initial enthalpy level,
- Drop by 802 kJ for the first intermediate reaction,
- Drop further by 88 kJ for the conversion of \(2 H_2O(g) \rightarrow 2 H_2O(l)\),
- End at a total enthalpy drop of 890 kJ from the starting point.
The enthalpy diagram should reflect these steps, forming a consistent path from reactants to products, with appropriate intermediate steps.
This detailed step-by-step process ensures that the enthalpy diagram aligns with Hess's Law for this chemical system.
