Question

Given: $ \Delta H_f^0 [C(graphite)] = 710 $ kJ mol⁻¹ $ \Delta_c H^0 = 414 $ kJ mol⁻¹ $ \Delta_{H-H}^0 = 436 $ kJ mol⁻¹ $ \Delta_{C-H}^0 = 611 $ kJ mol⁻¹   
The \(\Delta H_{C=C}^0 \text{ for }CH_2=CH_2 \text{ is }\) _____\(\text{ kJ mol}^{-1} \text{ (nearest integer value)}\)

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

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

Final Answer: The final answer is $25$