Question

1. calculate δh for the reaction: c₂h₄(g) + h₂(g) → c₂h₆(g) from the following:

(1) c₂h₄(g) + 3o₂(g) → 2co₂(g) + 2h₂o(l) δh = -1411 kj
(2) c₂h₆(g) + 7/2 o₂(g) → 2co₂(g) + 3h₂o(l) δh = -1560 kj
(3) h₂(g) + 1/2 o₂(g) → h₂o(l) δh = -285.8 kj

Explanation:

Step1: Define Target Reaction

Target: $\ce{C2H4(g) + H2(g) -> C2H6(g)}$

Step2: List Given Reactions

(1) $\ce{C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l)}$ $\Delta H_1 = -1411\ \text{kJ}$
(2) $\ce{C2H6(g) + 7/2O2(g) -> 2CO2(g) + 3H2O(l)}$ $\Delta H_2 = -1560\ \text{kJ}$
(3) $\ce{H2(g) + 1/2O2(g) -> H2O(l)}$ $\Delta H_3 = -285.8\ \text{kJ}$

Step3: Manipulate Reactions

• Reverse reaction (2): $\ce{2CO2(g) + 3H2O(l) -> C2H6(g) + 7/2O2(g)}$ $\Delta H_{2}' = +1560\ \text{kJ}$
• Keep reaction (1): $\ce{C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l)}$ $\Delta H_1 = -1411\ \text{kJ}$
• Keep reaction (3): $\ce{H2(g) + 1/2O2(g) -> H2O(l)}$ $\Delta H_3 = -285.8\ \text{kJ}$

Step4: Sum Manipulated Reactions

Add (1) + reversed (2) + (3):
$\ce{C2H4(g) + 3O2(g) + 2CO2(g) + 3H2O(l) + H2(g) + 1/2O2(g) -> 2CO2(g) + 2H2O(l) + C2H6(g) + 7/2O2(g) + H2O(l)}$

Simplify: $\ce{C2H4(g) + H2(g) -> C2H6(g)}$

Step5: Calculate $\Delta H$

$\Delta H = \Delta H_1 + \Delta H_{2}' + \Delta H_3$
$\Delta H = -1411 + 1560 - 285.8$
$\Delta H = -136.8\ \text{kJ}$

Answer:

$\boldsymbol{\Delta H = -136.8\ \text{kJ}}$ (or -137 kJ, depending on rounding)