Question

1. ethene, c₂h₄(g) is used in the manufacture of many polymers. if ethene could be formed from the elements carbon and hydrogen, the equation would be as follows: 2c(s) + 2h₂(g) → c₂h₄(g) use hess’s law and the equations given below to determine the molar enthalpy of formation for ethene. (1) c(s) + o₂(g) → co₂(g) δh° = -393.5 kj (2) h₂(g) + 1/2o₂(g) → h₂o(l) δh° = -285.8 kj (3) c₂h₄(g) + 3o₂(g) → 2co₂(g) + 2h₂o(l) δh° = -1411.2 kj

Explanation:

Step1: Define Target Reaction

The formation reaction of ethene (\(C_2H_4\)) is \(2C_{(s)} + 2H_{2(g)}
ightarrow C_2H_{4(g)}\). We need to find \(\Delta H_f^\circ\) for this reaction.

Step2: Analyze Given Reactions

1. Reaction (1): \(C_{(s)} + O_{2(g)}

ightarrow CO_{2(g)}\), \(\Delta H_1^\circ = -393.5\space kJ\)

1. Reaction (2): \(H_{2(g)} + \frac{1}{2}O_{2(g)}

ightarrow H_2O_{(l)}\), \(\Delta H_2^\circ = -285.8\space kJ\)

1. Reaction (3): \(C_2H_{4(g)} + 3O_{2(g)}

ightarrow 2CO_{2(g)} + 2H_2O_{(l)}\), \(\Delta H_3^\circ = -1411.2\space kJ\)

Step3: Manipulate Reactions

• Multiply Reaction (1) by 2: \(2C_{(s)} + 2O_{2(g)}

ightarrow 2CO_{2(g)}\), \(\Delta H_1'^\circ = 2\times(-393.5) = -787\space kJ\)

• Multiply Reaction (2) by 2: \(2H_{2(g)} + O_{2(g)}

ightarrow 2H_2O_{(l)}\), \(\Delta H_2'^\circ = 2\times(-285.8) = -571.6\space kJ\)

• Reverse Reaction (3): \(2CO_{2(g)} + 2H_2O_{(l)}

ightarrow C_2H_{4(g)} + 3O_{2(g)}\), \(\Delta H_3'^\circ = +1411.2\space kJ\)

Step4: Sum Manipulated Reactions

Add the three manipulated reactions:

$$
LATEXBLOCK0
$$

Simplify (cancel common terms): \(2C_{(s)} + 2H_{2(g)}
ightarrow C_2H_{4(g)}\)

Step5: Calculate \(\Delta H_f^\circ\)

Sum the enthalpies:

$$ \Delta H_f^\circ = \Delta H_1'^\circ + \Delta H_2'^\circ + \Delta H_3'^\circ $$

$$ \Delta H_f^\circ = -787 + (-571.6) + 1411.2 $$

$$ \Delta H_f^\circ = -787 - 571.6 + 1411.2 $$

$$ \Delta H_f^\circ = 52.6\space kJ $$

Answer:

The molar enthalpy of formation for ethene is \(\boldsymbol{+52.6\space kJ/mol}\)