Question

etylene, c₂h₄, reacts with hydrogen to produce methane, ch₄.
c₂h₄ (g) + 2 h₂ (g) → 2 ch₄ (g)
using hesss law, calculate the enthalpy change, δh°, for the above reaction:
2 c₂h₆ (g) + 7 o₂ (g) → 4 co₂ (g) + 6 h₂o (l); δh° = -3120.8 kj
ch₄ (g) + 2 o₂ (g) → co₂ (g) + 2 h₂o (l); δh° = -890.3 kj
c₂h₆ (g) → c₂h₄ (g) + h₂ (g); δh° = +136.3 kj
h₂ (g) + 1/2 o₂ (g); → h₂o (l); δh° = -285.8 kj
report your answer in kilojoules to the correct number of sig figs, but do not include units in your answer.

Explanation:

Step1: Identify Target Reaction

Target: $\ce{C2H4(g) + 2 H2(g) -> 2 CH4(g)}$

Step2: Manipulate Given Reactions

1. Reaction 1: $\ce{2 C2H6(g) + 7 O2(g) -> 4 CO2(g) + 6 H2O(l)}$; $\Delta H^\circ = -3120.8$ kJ
Divide by 2: $\ce{C2H6(g) + 3.5 O2(g) -> 2 CO2(g) + 3 H2O(l)}$; $\Delta H^\circ = -1560.4$ kJ

2. Reaction 2: $\ce{CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(l)}$; $\Delta H^\circ = -890.3$ kJ
Reverse and multiply by 2: $\ce{2 CO2(g) + 4 H2O(l) -> 2 CH4(g) + 4 O2(g)}$; $\Delta H^\circ = +1780.6$ kJ

3. Reaction 3: $\ce{C2H6(g) -> C2H4(g) + H2(g)}$; $\Delta H^\circ = +136.3$ kJ
Reverse: $\ce{C2H4(g) + H2(g) -> C2H6(g)}$; $\Delta H^\circ = -136.3$ kJ

4. Reaction 4: $\ce{H2(g) + 1/2 O2(g) -> H2O(l)}$; $\Delta H^\circ = -285.8$ kJ
Multiply by 1: $\ce{H2(g) + 1/2 O2(g) -> H2O(l)}$; $\Delta H^\circ = -285.8$ kJ (need 1 more $\ce{H2}$ reaction, so multiply by 1? Wait, target has 2 $\ce{H2}$. Wait, let's recheck.

Wait, let's sum the manipulated reactions:
- From Reaction 1 (divided): $\ce{C2H6 + 3.5 O2 -> 2 CO2 + 3 H2O}$; $\Delta H = -1560.4$
- From Reaction 2 (reversed ×2): $\ce{2 CO2 + 4 H2O -> 2 CH4 + 4 O2}$; $\Delta H = +1780.6$
- From Reaction 3 (reversed): $\ce{C2H4 + H2 -> C2H6}$; $\Delta H = -136.3$
- From Reaction 4 (×1): $\ce{H2 + 0.5 O2 -> H2O}$; $\Delta H = -285.8$ (need one more $\ce{H2}$? Wait, target has 2 $\ce{H2}$. Wait, let's add Reaction 4 again? Wait, no. Wait, let's check the $\ce{H2}$ count.

Wait, let's redo the manipulation:

After Reaction 1 (divided), Reaction 2 (reversed ×2), Reaction 3 (reversed), and Reaction 4 (×1) plus Reaction 4 (×1)? No, wait:

Wait, target has $\ce{C2H4 + 2 H2 -> 2 CH4}$. Let's track $\ce{C2H4}$, $\ce{H2}$, $\ce{CH4}$, $\ce{C2H6}$, $\ce{O2}$, $\ce{H2O}$, $\ce{CO2}$.

Sum the manipulated reactions:
1. $\ce{C2H6 + 3.5 O2 -> 2 CO2 + 3 H2O}$ ($\Delta H = -1560.4$)
2. $\ce{2 CO2 + 4 H2O -> 2 CH4 + 4 O2}$ ($\Delta H = +1780.6$)
3. $\ce{C2H4 + H2 -> C2H6}$ ($\Delta H = -136.3$)
4. $\ce{H2 + 0.5 O2 -> H2O}$ ($\Delta H = -285.8$)
5. Wait, need one more $\ce{H2}$ reaction? Wait, target has 2 $\ce{H2}$. In step 3, we have 1 $\ce{H2}$ (from reversed Reaction 3: $\ce{C2H4 + H2 -> C2H6}$), and we need 1 more $\ce{H2}$. So add Reaction 4 again? Wait, no. Wait, let's add Reaction 4 once more? Wait, no, let's check the $\ce{H2}$ in target: 2 $\ce{H2}$. Let's see:

After adding Reaction 3 reversed (1 $\ce{H2}$) and Reaction 4 (1 $\ce{H2}$), that's 2 $\ce{H2}$. Wait, Reaction 3 reversed: $\ce{C2H4 + H2 -> C2H6}$ (1 $\ce{H2}$), Reaction 4: $\ce{H2 + 0.5 O2 -> H2O}$ (1 $\ce{H2}$). So total 2 $\ce{H2}$. Good.

Now sum all manipulated reactions:
$\ce{C2H6 + 3.5 O2 + 2 CO2 + 4 H2O + C2H4 + H2 + H2 + 0.5 O2 -> 2 CO2 + 3 H2O + 2 CH4 + 4 O2 + C2H6 + H2O}$

Simplify:
- $\ce{C2H6}$ cancels.
- $\ce{2 CO2}$ cancels.
- $\ce{3.5 O2 + 0.5 O2 = 4 O2}$; $\ce{4 O2}$ cancels.
- $\ce{4 H2O + H2O = 5 H2O}$; $\ce{3 H2O + H2O = 4 H2O}$? Wait, no, let's do step-by-step:

Reactants: $\ce{C2H6, 3.5 O2, 2 CO2, 4 H2O, C2H4, H2, H2, 0.5 O2}$
Products: $\ce{2 CO2, 3 H2O, 2 CH4, 4 O2, C2H6, H2O}$

Cancel $\ce{C2H6}$:
Reactants: $\ce{3.5 O2, 2 CO2, 4 H2O, C2H4, H2, H2, 0.5 O2}$
Products: $\ce{2 CO2, 3 H2O, 2 CH4, 4 O2, H2O}$

Cancel $\ce{2 CO2}$:
Reactants: $\ce{4 O2 (3.5 + 0.5), 4 H2O, C2H4, 2 H2}$
Products: $\ce{4 H2O (3 + 1), 2 CH4, 4 O2}$

Cancel $\ce{4 O2}$ and $\ce{4 H2O}$:
Reactants: $\ce{C2H4, 2 H2}$
Products: $\ce{2 CH4}$

Perfect! Now sum the $\Delta H$:
-1560.4 (Reaction1/2) + 1780.6 (Reaction2 reversed×2) -136.3 (Reaction3 reversed) -285.8 (Reaction4) -285.8? Wait no, wait Reaction4 was multiplied by 1? Wait, no, in the manipulated reactions, we had Reaction4 as $\ce{H2 + 0.5 O2 -> H2O}$ (1 $\ce{H2}$), but we needed 2 $\ce{H2}$ in target. Wait, no, in the reversed Reaction3, we have 1 $\ce{H2}$, and then we need 1 more $\ce{H2}$, so we should multiply Reaction4 by 1 (for 1 $\ce{H2}$) and add another Reaction4? Wait, no, let's recalculate the $\Delta H$:

Wait, the correct manipulation:

We have:
1. Reaction1/2: $\Delta H = -1560.4$
2. Reaction2 reversed×2: $\Delta H = +1780.6$
3. Reaction3 reversed: $\Delta H = -136.3$
4. Reaction4 (for 1 $\ce{H2}$): $\Delta H = -285.8$
5. Reaction4 (for the second $\ce{H2}$): $\Delta H = -285.8$ (wait, no, target has 2 $\ce{H2}$. Wait, in the target, after Reaction3 reversed (1 $\ce{H2}$), we need 1 more $\ce{H2}$, so we use Reaction4 once more. Wait, no, let's check the sum again.

Wait, let's list all manipulated $\Delta H$:
- Reaction1/2: -1560.4
- Reaction2 reversed×2: +1780.6
- Reaction3 reversed: -136.3
- Reaction4 (×1): -285.8 (for 1 $\ce{H2}$)
- Reaction4 (×1): -285.8 (for the second $\ce{H2}$)

Wait, no, that's incorrect. Wait, in the target, the $\ce{H2}$ comes from Reaction3 reversed (1 $\ce{H2}$) and Reaction4 (1 $\ce{H2}$), so total 2 $\ce{H2}$. So Reaction4 is used once. Wait, no, in the simplified reaction, we had 2 $\ce{H2}$ in reactants, which came from Reaction3 reversed (1 $\ce{H2}$) and Reaction4 (1 $\ce{H2}$). So Reaction4 is used once. Wait, but in the $\Delta H$ sum, let's recalculate:

-1560.4 (Reaction1/2) + 1780.6 (Reaction2 reversed×2) -136.3 (Reaction3 reversed) -285.8 (Reaction4) = ?

Calculate:
-1560.4 + 1780.6 = 220.2
220.2 - 136.3 = 83.9
83.9 - 285.8 = -201.9

Wait, that can't be. Wait, no, I missed a Reaction4. Wait, target has 2 $\ce{H2}$, but Reaction3 reversed gives 1 $\ce{H2}$, so we need 1 more $\ce{H2}$, which comes from Reaction4? No, Reaction4 is $\ce{H2 + 0.5 O2 -> H2O}$, so if we use Reaction4 once, that's 1 $\ce{H2}$, but we need 2 $\ce{H2}$ in target. Wait, no, let's re-express the target:

Target: $\ce{C2H4 + 2 H2 -> 2 CH4}$

From Reaction3 reversed: $\ce{C2H4 + H2 -> C2H6}$ (1 $\ce{H2}$)
From Reaction4: $\ce{H2 + 0.5 O2 -> H2O}$ (1 $\ce{H2}$)
Then, from Reaction1/2: $\ce{C2H6 + 3.5 O2 -> 2 CO2 + 3 H2O}$
From Reaction2 reversed×2: $\ce{2 CO2 + 4 H2O -> 2 CH4 + 4 O2}$

Now, sum these four reactions:

1. $\ce{C2H4 + H2 -> C2H6}$ (ΔH -136.3)
2. $\ce{C2H6 + 3.5 O2 -> 2 CO2 + 3 H2O}$ (ΔH -1560.4)
3. $\ce{2 CO2 + 4 H2O -> 2 CH4 + 4 O2}$ (ΔH +1780.6)
4. $\ce{H2 + 0.5 O2 -> H2O}$ (ΔH -285.8)

Now sum these four:

Reactants: $\ce{C2H4 + H2 + C2H6 + 3.5 O2 + 2 CO2 + 4 H2O + H2 + 0.5 O2}$
Products: $\ce{C2H6 + 2 CO2 + 3 H2O + 2 CH4 + 4 O2 + H2O}$

Simplify:
- $\ce{C2H6}$ cancels.
- $\ce{2 CO2}$ cancels.
- $\ce{3.5 O2 + 0.5 O2 = 4 O2}$; $\ce{4 O2}$ cancels.
- $\ce{4 H2O + H2O = 5 H2O}$; $\ce{3 H2O + H2O = 4 H2O}$? No, 3 H2O (from Reaction2) + 1 H2O (from Reaction4) = 4 H2O; 4 H2O (from Reaction3 reversed×2) cancels with 4 H2O (from Reaction2 + Reaction4).
- $\ce{H2 + H2 = 2 H2}$ (reactants)
- $\ce{C2H4}$ (reactant)
- $\ce{2 CH4}$ (product)

Perfect! Now sum ΔH:
-136.3 (Reaction3 reversed) + (-1560.4) (Reaction1/2) + 1780.6 (Reaction2 reversed×2) + (-285.8) (Reaction4)

Calculate:
-136.3 -1560.4 = -1696.7
-1696.7 + 1780.6 = 83.9
83.9 - 285.8 = -201.9

Wait, but let's check with another approach. Let's write the target reaction and express it as a combination of given reactions.

Given reactions:
R1: $\ce{2 C2H6 + 7 O2 -> 4 CO2 + 6 H2O}$; ΔH = -3120.8
R2: $\ce{CH4 + 2 O2 -> CO2 + 2 H2O}$; ΔH = -890.3
R3: $\ce{C2H6 -> C2H4 + H2}$; ΔH = +136.3
R4: $\ce{H2 + 0.5 O2 -> H2O}$; ΔH = -285.8

Target: $\ce{C2H4 + 2 H2 -> 2 CH4}$

Let’s express target as:
Target = -R3 + (R1/2) - 2×R2 + R4

Wait, let's check:

-R3: $\ce{C2H4 + H2 -> C2H6}$; ΔH = -136.3
R1/2: $\ce{C2H6 + 3.5 O2 -> 2 CO2 + 3 H2O}$; ΔH = -1560.4
-2×R2: $\ce{2 CO2 + 4 H2O -> 2 CH4 + 4 O2}$; ΔH = +1780.6
R4: $\ce{H2 + 0.5 O2 -> H2O}$; ΔH = -285.8

Sum these:
$\ce{C2H4 + H2 + C2H6 + 3.5 O2 + 2 CO2 + 4 H2O + H2 + 0.5 O2 -> C2H6 + 2 CO2 + 3 H2O + 2 CH4 + 4 O2 + H2O}$

Simplify to $\ce{C2H4 + 2 H2 -> 2 CH4}$, correct.

Now sum ΔH:
-136.3 + (-1560.4) + 1780.6 + (-285.8) = (-136.3 -1560.4 -285.8) + 1780.6 = (-1982.5) + 1780.6 = -201.9

Yes, that's correct. So the enthalpy change is -202 (rounded to three significant figures? Wait, given values: -3120.8 (5 sig figs), -890.3 (4), +136.3 (4), -285.8 (4). The least number of decimal places? No, sig figs: 136.3 is 4, 285.8 is 4, 890.3 is 4, 3120.8 is 5. So the answer should have 4 sig figs? Wait, -201.9 rounds to -202 (three sig figs? Wait, 201.9 is four sig figs). Wait, let's check the calculation again.

Wait, 1780.6 - 1560.4 = 220.2; 220.2 - 136.3 = 83.9; 83.9 - 285.8 = -201.9. So -201.9, which is -202 when rounded to three significant figures? Wait, 201.9 has four significant figures. The given ΔH values:

Answer:

Step1: Identify Target Reaction

Target: $\ce{C2H4(g) + 2 H2(g) -> 2 CH4(g)}$

Step2: Manipulate Given Reactions

1. Reaction 1: $\ce{2 C2H6(g) + 7 O2(g) -> 4 CO2(g) + 6 H2O(l)}$; $\Delta H^\circ = -3120.8$ kJ

Divide by 2: $\ce{C2H6(g) + 3.5 O2(g) -> 2 CO2(g) + 3 H2O(l)}$; $\Delta H^\circ = -1560.4$ kJ

1. Reaction 2: $\ce{CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(l)}$; $\Delta H^\circ = -890.3$ kJ

Reverse and multiply by 2: $\ce{2 CO2(g) + 4 H2O(l) -> 2 CH4(g) + 4 O2(g)}$; $\Delta H^\circ = +1780.6$ kJ

1. Reaction 3: $\ce{C2H6(g) -> C2H4(g) + H2(g)}$; $\Delta H^\circ = +136.3$ kJ

Reverse: $\ce{C2H4(g) + H2(g) -> C2H6(g)}$; $\Delta H^\circ = -136.3$ kJ

1. Reaction 4: $\ce{H2(g) + 1/2 O2(g) -> H2O(l)}$; $\Delta H^\circ = -285.8$ kJ

Multiply by 1: $\ce{H2(g) + 1/2 O2(g) -> H2O(l)}$; $\Delta H^\circ = -285.8$ kJ (need 1 more $\ce{H2}$ reaction, so multiply by 1? Wait, target has 2 $\ce{H2}$. Wait, let's recheck.

Wait, let's sum the manipulated reactions:

• From Reaction 1 (divided): $\ce{C2H6 + 3.5 O2 -> 2 CO2 + 3 H2O}$; $\Delta H = -1560.4$
• From Reaction 2 (reversed ×2): $\ce{2 CO2 + 4 H2O -> 2 CH4 + 4 O2}$; $\Delta H = +1780.6$
• From Reaction 3 (reversed): $\ce{C2H4 + H2 -> C2H6}$; $\Delta H = -136.3$
• From Reaction 4 (×1): $\ce{H2 + 0.5 O2 -> H2O}$; $\Delta H = -285.8$ (need one more $\ce{H2}$? Wait, target has 2 $\ce{H2}$. Wait, let's add Reaction 4 again? Wait, no. Wait, let's check the $\ce{H2}$ count.

Wait, let's redo the manipulation:

After Reaction 1 (divided), Reaction 2 (reversed ×2), Reaction 3 (reversed), and Reaction 4 (×1) plus Reaction 4 (×1)? No, wait:

Wait, target has $\ce{C2H4 + 2 H2 -> 2 CH4}$. Let's track $\ce{C2H4}$, $\ce{H2}$, $\ce{CH4}$, $\ce{C2H6}$, $\ce{O2}$, $\ce{H2O}$, $\ce{CO2}$.

Sum the manipulated reactions:

1. $\ce{C2H6 + 3.5 O2 -> 2 CO2 + 3 H2O}$ ($\Delta H = -1560.4$)
2. $\ce{2 CO2 + 4 H2O -> 2 CH4 + 4 O2}$ ($\Delta H = +1780.6$)
3. $\ce{C2H4 + H2 -> C2H6}$ ($\Delta H = -136.3$)
4. $\ce{H2 + 0.5 O2 -> H2O}$ ($\Delta H = -285.8$)
5. Wait, need one more $\ce{H2}$ reaction? Wait, target has 2 $\ce{H2}$. In step 3, we have 1 $\ce{H2}$ (from reversed Reaction 3: $\ce{C2H4 + H2 -> C2H6}$), and we need 1 more $\ce{H2}$. So add Reaction 4 again? Wait, no. Wait, let's add Reaction 4 once more? Wait, no, let's check the $\ce{H2}$ in target: 2 $\ce{H2}$. Let's see:

After adding Reaction 3 reversed (1 $\ce{H2}$) and Reaction 4 (1 $\ce{H2}$), that's 2 $\ce{H2}$. Wait, Reaction 3 reversed: $\ce{C2H4 + H2 -> C2H6}$ (1 $\ce{H2}$), Reaction 4: $\ce{H2 + 0.5 O2 -> H2O}$ (1 $\ce{H2}$). So total 2 $\ce{H2}$. Good.

Now sum all manipulated reactions:
$\ce{C2H6 + 3.5 O2 + 2 CO2 + 4 H2O + C2H4 + H2 + H2 + 0.5 O2 -> 2 CO2 + 3 H2O + 2 CH4 + 4 O2 + C2H6 + H2O}$

Simplify:

• $\ce{C2H6}$ cancels.
• $\ce{2 CO2}$ cancels.
• $\ce{3.5 O2 + 0.5 O2 = 4 O2}$; $\ce{4 O2}$ cancels.
• $\ce{4 H2O + H2O = 5 H2O}$; $\ce{3 H2O + H2O = 4 H2O}$? Wait, no, let's do step-by-step:

Reactants: $\ce{C2H6, 3.5 O2, 2 CO2, 4 H2O, C2H4, H2, H2, 0.5 O2}$
Products: $\ce{2 CO2, 3 H2O, 2 CH4, 4 O2, C2H6, H2O}$

Cancel $\ce{C2H6}$:
Reactants: $\ce{3.5 O2, 2 CO2, 4 H2O, C2H4, H2, H2, 0.5 O2}$
Products: $\ce{2 CO2, 3 H2O, 2 CH4, 4 O2, H2O}$

Cancel $\ce{2 CO2}$:
Reactants: $\ce{4 O2 (3.5 + 0.5), 4 H2O, C2H4, 2 H2}$
Products: $\ce{4 H2O (3 + 1), 2 CH4, 4 O2}$

Cancel $\ce{4 O2}$ and $\ce{4 H2O}$:
Reactants: $\ce{C2H4, 2 H2}$
Products: $\ce{2 CH4}$

Perfect! Now sum the $\Delta H$:
-1560.4 (Reaction1/2) + 1780.6 (Reaction2 reversed×2) -136.3 (Reaction3 reversed) -285.8 (Reaction4) -285.8? Wait no, wait Reaction4 was multiplied by 1? Wait, no, in the manipulated reactions, we had Reaction4 as $\ce{H2 + 0.5 O2 -> H2O}$ (1 $\ce{H2}$), but we needed 2 $\ce{H2}$ in target. Wait, no, in the reversed Reaction3, we have 1 $\ce{H2}$, and then we need 1 more $\ce{H2}$, so we should multiply Reaction4 by 1 (for 1 $\ce{H2}$) and add another Reaction4? Wait, no, let's recalculate the $\Delta H$:

Wait, the correct manipulation:

We have:

1. Reaction1/2: $\Delta H = -1560.4$
2. Reaction2 reversed×2: $\Delta H = +1780.6$
3. Reaction3 reversed: $\Delta H = -136.3$
4. Reaction4 (for 1 $\ce{H2}$): $\Delta H = -285.8$
5. Reaction4 (for the second $\ce{H2}$): $\Delta H = -285.8$ (wait, no, target has 2 $\ce{H2}$. Wait, in the target, after Reaction3 reversed (1 $\ce{H2}$), we need 1 more $\ce{H2}$, so we use Reaction4 once more. Wait, no, let's check the sum again.

Wait, let's list all manipulated $\Delta H$:

• Reaction1/2: -1560.4
• Reaction2 reversed×2: +1780.6
• Reaction3 reversed: -136.3
• Reaction4 (×1): -285.8 (for 1 $\ce{H2}$)
• Reaction4 (×1): -285.8 (for the second $\ce{H2}$)

Wait, no, that's incorrect. Wait, in the target, the $\ce{H2}$ comes from Reaction3 reversed (1 $\ce{H2}$) and Reaction4 (1 $\ce{H2}$), so total 2 $\ce{H2}$. So Reaction4 is used once. Wait, no, in the simplified reaction, we had 2 $\ce{H2}$ in reactants, which came from Reaction3 reversed (1 $\ce{H2}$) and Reaction4 (1 $\ce{H2}$). So Reaction4 is used once. Wait, but in the $\Delta H$ sum, let's recalculate:

-1560.4 (Reaction1/2) + 1780.6 (Reaction2 reversed×2) -136.3 (Reaction3 reversed) -285.8 (Reaction4) = ?

Calculate:
-1560.4 + 1780.6 = 220.2
220.2 - 136.3 = 83.9
83.9 - 285.8 = -201.9

Wait, that can't be. Wait, no, I missed a Reaction4. Wait, target has 2 $\ce{H2}$, but Reaction3 reversed gives 1 $\ce{H2}$, so we need 1 more $\ce{H2}$, which comes from Reaction4? No, Reaction4 is $\ce{H2 + 0.5 O2 -> H2O}$, so if we use Reaction4 once, that's 1 $\ce{H2}$, but we need 2 $\ce{H2}$ in target. Wait, no, let's re-express the target:

Target: $\ce{C2H4 + 2 H2 -> 2 CH4}$

From Reaction3 reversed: $\ce{C2H4 + H2 -> C2H6}$ (1 $\ce{H2}$)
From Reaction4: $\ce{H2 + 0.5 O2 -> H2O}$ (1 $\ce{H2}$)
Then, from Reaction1/2: $\ce{C2H6 + 3.5 O2 -> 2 CO2 + 3 H2O}$
From Reaction2 reversed×2: $\ce{2 CO2 + 4 H2O -> 2 CH4 + 4 O2}$

Now, sum these four reactions:

1. $\ce{C2H4 + H2 -> C2H6}$ (ΔH -136.3)
2. $\ce{C2H6 + 3.5 O2 -> 2 CO2 + 3 H2O}$ (ΔH -1560.4)
3. $\ce{2 CO2 + 4 H2O -> 2 CH4 + 4 O2}$ (ΔH +1780.6)
4. $\ce{H2 + 0.5 O2 -> H2O}$ (ΔH -285.8)

Now sum these four:

Reactants: $\ce{C2H4 + H2 + C2H6 + 3.5 O2 + 2 CO2 + 4 H2O + H2 + 0.5 O2}$
Products: $\ce{C2H6 + 2 CO2 + 3 H2O + 2 CH4 + 4 O2 + H2O}$

Simplify:

• $\ce{C2H6}$ cancels.
• $\ce{2 CO2}$ cancels.
• $\ce{3.5 O2 + 0.5 O2 = 4 O2}$; $\ce{4 O2}$ cancels.
• $\ce{4 H2O + H2O = 5 H2O}$; $\ce{3 H2O + H2O = 4 H2O}$? No, 3 H2O (from Reaction2) + 1 H2O (from Reaction4) = 4 H2O; 4 H2O (from Reaction3 reversed×2) cancels with 4 H2O (from Reaction2 + Reaction4).
• $\ce{H2 + H2 = 2 H2}$ (reactants)
• $\ce{C2H4}$ (reactant)
• $\ce{2 CH4}$ (product)

Perfect! Now sum ΔH:
-136.3 (Reaction3 reversed) + (-1560.4) (Reaction1/2) + 1780.6 (Reaction2 reversed×2) + (-285.8) (Reaction4)

Calculate:
-136.3 -1560.4 = -1696.7
-1696.7 + 1780.6 = 83.9
83.9 - 285.8 = -201.9

Wait, but let's check with another approach. Let's write the target reaction and express it as a combination of given reactions.

Given reactions:
R1: $\ce{2 C2H6 + 7 O2 -> 4 CO2 + 6 H2O}$; ΔH = -3120.8
R2: $\ce{CH4 + 2 O2 -> CO2 + 2 H2O}$; ΔH = -890.3
R3: $\ce{C2H6 -> C2H4 + H2}$; ΔH = +136.3
R4: $\ce{H2 + 0.5 O2 -> H2O}$; ΔH = -285.8

Target: $\ce{C2H4 + 2 H2 -> 2 CH4}$

Let’s express target as:
Target = -R3 + (R1/2) - 2×R2 + R4

Wait, let's check:

-R3: $\ce{C2H4 + H2 -> C2H6}$; ΔH = -136.3
R1/2: $\ce{C2H6 + 3.5 O2 -> 2 CO2 + 3 H2O}$; ΔH = -1560.4
-2×R2: $\ce{2 CO2 + 4 H2O -> 2 CH4 + 4 O2}$; ΔH = +1780.6
R4: $\ce{H2 + 0.5 O2 -> H2O}$; ΔH = -285.8

Sum these:
$\ce{C2H4 + H2 + C2H6 + 3.5 O2 + 2 CO2 + 4 H2O + H2 + 0.5 O2 -> C2H6 + 2 CO2 + 3 H2O + 2 CH4 + 4 O2 + H2O}$

Simplify to $\ce{C2H4 + 2 H2 -> 2 CH4}$, correct.

Now sum ΔH:
-136.3 + (-1560.4) + 1780.6 + (-285.8) = (-136.3 -1560.4 -285.8) + 1780.6 = (-1982.5) + 1780.6 = -201.9

Yes, that's correct. So the enthalpy change is -202 (rounded to three significant figures? Wait, given values: -3120.8 (5 sig figs), -890.3 (4), +136.3 (4), -285.8 (4). The least number of decimal places? No, sig figs: 136.3 is 4, 285.8 is 4, 890.3 is 4, 3120.8 is 5. So the answer should have 4 sig figs? Wait, -201.9 rounds to -202 (three sig figs? Wait, 201.9 is four sig figs). Wait, let's check the calculation again.

Wait, 1780.6 - 1560.4 = 220.2; 220.2 - 136.3 = 83.9; 83.9 - 285.8 = -201.9. So -201.9, which is -202 when rounded to three significant figures? Wait, 201.9 has four significant figures. The given ΔH values: