Question

Given:

\[ \Delta^{\Theta}_{sub}[\text{C(graphite)}] = 710 \, kJ \, mol^{-1} \] \[ \Delta_{\text{C-H}}^{\Theta} = 414 \, kJ \, mol^{-1} \] \[ \Delta_{\text{H-H}}^{\Theta} = 436 \, kJ \, mol^{-1} \] \[ \Delta_{\text{C=C}}^{\Theta} = 611 \, kJ \, mol^{-1} \]

The \(\Delta H_f^{\Theta}\) for \(CH_2 = CH_2\) is _______ \(kJ \, mol^{-1}\) (nearest integer value).

Hint:

The enthalpy of formation of a compound can be related to the enthalpies of formation of its constituent atoms and the bond enthalpies of the bonds within the molecule. Set up an equation relating the enthalpy of formation of ethylene to the sublimation enthalpy of carbon, the bond enthalpy of hydrogen, and the bond enthalpies of C=C and C-H bonds, then solve for the C=C bond enthalpy.

Answer 1

Correct Answer: 25

We are given thermochemical data for the formation of ethene (\( CH_2 = CH_2 \)) and need to calculate its standard enthalpy of formation (\( \Delta H_f^\Theta \)).

Concept Used:

The standard enthalpy of formation (\( \Delta H_f^\Theta \)) of a compound is the enthalpy change when one mole of the compound is formed from its elements in their standard states. It can be calculated using bond energies and atomization energies:

\[ \Delta H_f^\Theta = \sum (\text{Bond energies of products}) - \sum (\text{Bond energies of reactants}) \]

For ethene (\( CH_2 = CH_2 \)), the reaction is:

\[ 2C(\text{graphite}) + 2H_2(g) \rightarrow C_2H_4(g) \]

Step-by-Step Solution:

Step 1: Write down the given data.

\[ \Delta_\text{sub} [C(\text{graphite})] = 710~kJ~mol^{-1} \] \[ \Delta_{C-H} = 414~kJ~mol^{-1} \] \[ \Delta_{H-H} = 436~kJ~mol^{-1} \] \[ \Delta_{C=C} = 611~kJ~mol^{-1} \]

Step 2: Determine the bonds in ethene (\( C_2H_4 \)).

In ethene, there are:

• 4 C–H bonds
• 1 C=C bond

Step 3: Calculate the energy required to atomize the reactants.

For the formation of gaseous atoms from elements in their standard states:

\[ 2C(\text{graphite}) \rightarrow 2C(g) \quad \text{requires } 2 \times 710 = 1420~kJ \] \[ 2H_2(g) \rightarrow 4H(g) \quad \text{requires } 2 \times 436 = 872~kJ \] \[ \text{Total energy to atomize reactants} = 1420 + 872 = 2292~kJ \]

Step 4: Calculate the energy released in forming bonds of \( C_2H_4 \).

\[ 4(C-H) + 1(C=C) \] \[ = 4(414) + 611 = 1656 + 611 = 2267~kJ \]

Step 5: Compute the enthalpy of formation.

\[ \Delta H_f^\Theta = \text{Energy required} - \text{Energy released} \] \[ \Delta H_f^\Theta = 2292 - 2267 = 25~kJ~mol^{-1} \]

Step 6: Since the formation of bonds releases energy, the enthalpy of formation is exothermic:

\[ \Delta H_f^\Theta = -25~kJ~mol^{-1} \]

Final Answer: The standard enthalpy of formation of \( CH_2 = CH_2 \) is 25 kJ mol⁻¹.

Answer 2

$[\Delta H_f^0]_{C_2H_4(g)} = (2 \times 710) + (2 \times 436) - 611 - 4 \times 414$
$[\Delta H_f^0]_{C_2H_4(g)} = 1420 + 872 - 611 - 1656$
$[\Delta H_f^0]_{C_2H_4(g)} = 2292 - 2267 = 25 \text{ kJ mol}^{-1}$

Final Answer: The final answer is $25$