Answer :
To solve the problem and construct the correct enthalpy diagram using Hess's law, we need to analyze the given intermediate reactions and see how we can derive the overall reaction from them.
Step 1: Write out the intermediate reactions with their respective enthalpies.
1. [tex]\( \text{Reaction 1:} \quad CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g) \quad \Delta H_1 = -802 \, \text{kJ} \)[/tex]
2. [tex]\( \text{Reaction 2:} \quad 2 H_2O(g) \rightarrow 2 H_2O(l) \quad \Delta H_2 = -88 \, \text{kJ} \)[/tex]
Step 2: Determine the overall reaction we are trying to achieve.
The overall chemical reaction is:
[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \][/tex]
Step 3: Apply Hess's law.
According to Hess's law, the total enthalpy change of a reaction is the sum of the enthalpy changes of the intermediate steps. Therefore, we sum the given intermediate enthalpies to obtain the total enthalpy change for the overall reaction.
[tex]\[ \Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 \][/tex]
Step 4: Substitute the values of [tex]\(\Delta H_1\)[/tex] and [tex]\(\Delta H_2\)[/tex].
[tex]\[ \Delta H_{\text{total}} = -802 \, \text{kJ} + (-88 \, \text{kJ}) = -890 \, \text{kJ} \][/tex]
Step 5: Construct the enthalpy diagram.
To complete the enthalpy diagram, we represent the given reactions with their respective enthalpy changes along the y-axis, plotting the initial, intermediate, and final states.
1. Start with [tex]\(CH_4(g) + 2 O_2(g)\)[/tex] at the initial energy level.
2. Move to [tex]\(CO_2(g) + 2 H_2O(g)\)[/tex] at a lower energy level, given by [tex]\(\Delta H_1 = -802 \, \text{kJ}\)[/tex].
3. Transition from [tex]\(CO_2(g) + 2 H_2O(g)\)[/tex] to [tex]\(CO_2(g) + 2 H_2O(l)\)[/tex], which further lowers the energy level by [tex]\(\Delta H_2 = -88 \, \text{kJ}\)[/tex].
4. The total enthalpy change from the initial state [tex]\(CH_4(g) + 2 O_2(g)\)[/tex] to the final state [tex]\(CO_2(g) + 2 H_2O(l)\)[/tex] is [tex]\(\Delta H_{\text{total}} = -890 \, \text{kJ}\)[/tex].
The enthalpy diagram should depict:
- A starting point labeled [tex]\(CH_4(g) + 2 O_2(g)\)[/tex].
- A downward arrow to an intermediate point labeled [tex]\(CO_2(g) + 2 H_2O(g)\)[/tex] with [tex]\(\Delta H_1 = -802 \, \text{kJ}\)[/tex].
- A further downward arrow to the final point labeled [tex]\(CO_2(g) + 2 H_2O(l)\)[/tex] with an additional [tex]\(\Delta H_2 = -88 \, \text{kJ}\)[/tex].
- The total enthalpy change of [tex]\(\Delta H_{\text{total}} = -890 \, \text{kJ}\)[/tex] shown from the initial state to the final state.
By following these steps, you will have a clear representation of the enthalpy changes using Hess's law for the given system.
Step 1: Write out the intermediate reactions with their respective enthalpies.
1. [tex]\( \text{Reaction 1:} \quad CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g) \quad \Delta H_1 = -802 \, \text{kJ} \)[/tex]
2. [tex]\( \text{Reaction 2:} \quad 2 H_2O(g) \rightarrow 2 H_2O(l) \quad \Delta H_2 = -88 \, \text{kJ} \)[/tex]
Step 2: Determine the overall reaction we are trying to achieve.
The overall chemical reaction is:
[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \][/tex]
Step 3: Apply Hess's law.
According to Hess's law, the total enthalpy change of a reaction is the sum of the enthalpy changes of the intermediate steps. Therefore, we sum the given intermediate enthalpies to obtain the total enthalpy change for the overall reaction.
[tex]\[ \Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 \][/tex]
Step 4: Substitute the values of [tex]\(\Delta H_1\)[/tex] and [tex]\(\Delta H_2\)[/tex].
[tex]\[ \Delta H_{\text{total}} = -802 \, \text{kJ} + (-88 \, \text{kJ}) = -890 \, \text{kJ} \][/tex]
Step 5: Construct the enthalpy diagram.
To complete the enthalpy diagram, we represent the given reactions with their respective enthalpy changes along the y-axis, plotting the initial, intermediate, and final states.
1. Start with [tex]\(CH_4(g) + 2 O_2(g)\)[/tex] at the initial energy level.
2. Move to [tex]\(CO_2(g) + 2 H_2O(g)\)[/tex] at a lower energy level, given by [tex]\(\Delta H_1 = -802 \, \text{kJ}\)[/tex].
3. Transition from [tex]\(CO_2(g) + 2 H_2O(g)\)[/tex] to [tex]\(CO_2(g) + 2 H_2O(l)\)[/tex], which further lowers the energy level by [tex]\(\Delta H_2 = -88 \, \text{kJ}\)[/tex].
4. The total enthalpy change from the initial state [tex]\(CH_4(g) + 2 O_2(g)\)[/tex] to the final state [tex]\(CO_2(g) + 2 H_2O(l)\)[/tex] is [tex]\(\Delta H_{\text{total}} = -890 \, \text{kJ}\)[/tex].
The enthalpy diagram should depict:
- A starting point labeled [tex]\(CH_4(g) + 2 O_2(g)\)[/tex].
- A downward arrow to an intermediate point labeled [tex]\(CO_2(g) + 2 H_2O(g)\)[/tex] with [tex]\(\Delta H_1 = -802 \, \text{kJ}\)[/tex].
- A further downward arrow to the final point labeled [tex]\(CO_2(g) + 2 H_2O(l)\)[/tex] with an additional [tex]\(\Delta H_2 = -88 \, \text{kJ}\)[/tex].
- The total enthalpy change of [tex]\(\Delta H_{\text{total}} = -890 \, \text{kJ}\)[/tex] shown from the initial state to the final state.
By following these steps, you will have a clear representation of the enthalpy changes using Hess's law for the given system.
