Step 1: Understanding the Concept:
We need to calculate the enthalpy of the target reaction using Hess's Law. We manipulate the given thermochemical equations to sum up to the target reaction.
Step 2: Key Formula or Approach:
Target: \( \text{CH}_4 \to \text{C} + 2\text{H}_2 \) (Wait, let's balance the given target).
The given target equation in the question is: \( \text{CH}_4(g) + \text{O}_2(g) \to \text{C}(s) + 2\text{H}_2\text{O}(l) \).
Wait, stoichiometry:
C: 1 \(\to\) 1.
H: 4 \(\to\) 4 (in 2 H2O).
O: 2 \(\to\) 2 (in 2 H2O).
Balanced.
Step 3: Detailed Explanation:
Let's label the given equations:
1. \( \text{H}_2 + 0.5\text{O}_2 \to \text{H}_2\text{O} \), \( \Delta H_1 = -286 \).
2. \( \text{C} + \text{O}_2 \to \text{CO}_2 \), \( \Delta H_2 = -394 \).
3. \( \text{CH}_4 + 2\text{O}_2 \to \text{CO}_2 + 2\text{H}_2\text{O} \), \( \Delta H_3 = -890 \).
We want: \( \text{CH}_4 + \text{O}_2 \to \text{C} + 2\text{H}_2\text{O} \).
Look at the components:
- \( \text{CH}_4 \) is on the LHS. Equation 3 has \( \text{CH}_4 \) on LHS. Keep Eq 3.
- \( \text{C} \) is on the RHS. Equation 2 has \( \text{C} \) on LHS. Reverse Eq 2.
- \( \text{H}_2\text{O} \) is on the RHS. Eq 3 has 2 \( \text{H}_2\text{O} \) on RHS. This matches.
- Check \( \text{CO}_2 \): Eq 3 produces 1 \( \text{CO}_2 \). Reverse Eq 2 consumes 1 \( \text{CO}_2 \). They cancel.
- Check \( \text{O}_2 \): Eq 3 uses 2 \( \text{O}_2 \). Reverse Eq 2 produces 1 \( \text{O}_2 \). Net: \( 2 - 1 = 1 \) \( \text{O}_2 \) on LHS. This matches.
Operation: Equation 3 - Equation 2.
\( (\text{CH}_4 + 2\text{O}_2) - (\text{C} + \text{O}_2) \to (\text{CO}_2 + 2\text{H}_2\text{O}) - (\text{CO}_2) \)
\( \text{CH}_4 + \text{O}_2 \to \text{C} + 2\text{H}_2\text{O} \).
This matches the target exactly. Eq 1 is not needed.
Calculation:
\( \Delta H = \Delta H_3 - \Delta H_2 \)
\( \Delta H = -890 - (-394) \)
\( \Delta H = -890 + 394 \)
\( \Delta H = -496 \text{ kJ} \).
Wait, the options include +496 (A) and -496 (B).
Let's check the Answer Key. The green check is on Option 1 (+496).
Why?
Let me re-read the target reaction.
\( \text{CH}_4(g) + \text{O}_2(g) \to \text{C}(s) + 2\text{H}_2\text{O}(l) \).
My derivation gives -496.
Is it possible the target is different?
Maybe \( \text{C}(s) + 2\text{H}_2\text{O} \to \text{CH}_4 + \text{O}_2 \)? No, combustion related.
Let's assume the question asks for the enthalpy of *formation* or something else? No, "enthalpy change for the reaction".
Let's re-calculate \( \Delta H_3 - \Delta H_2 \).
-890 + 394 = -496.
Is there a sign error in my data reading?
Data: -286, -394, -890. All standard combustion enthalpies.
Target: Incomplete combustion of Methane to Carbon (soot).
Combustion is exothermic. -496 makes physical sense. +496 would be endothermic.
However, if the key says +496, maybe the reaction is reversed?
Reaction: \( \text{CH}_4 + \text{O}_2 \to \text{C} + 2\text{H}_2\text{O} \).
This is oxidation.
Let's look at the screenshot again.
Image 4, Question 129.
Option 1: +496. (Red cross).
Option 2: -496. (Green tick).
Ah! The green check mark is on Option 2. The red cross is on Option 1.
I misread the visual cues in the prompt or confused the standard format. In the provided images, the green tick indicates the correct answer.
My calculation (-496) matches Option 2.
Step 4: Final Answer:
The enthalpy change is -496 kJ.