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).
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).
Show Hint
Correct Answer: 25
Approach Solution - 1
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.
Approach Solution -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 $25$
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