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 \(\Delta H_{C=C}^0 \text{ for }CH_2=CH_2 \text{ is }\) _____\(\text{ kJ mol}^{-1} \text{ (nearest integer value)}\)
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 \(\Delta H_{C=C}^0 \text{ for }CH_2=CH_2 \text{ is }\) _____\(\text{ kJ mol}^{-1} \text{ (nearest integer value)}\)
Show Hint
Correct Answer: 25
Solution and Explanation
$[\Delta H_f^0]_{C_2H_4(g)} = (2 \times 710) + (2 \times 436) - 611 - 4 \times 414$
$[\Delta H_f^0]_{C_2H_4(g)} = 1420 + 872 - 611 - 1656$
$[\Delta H_f^0]_{C_2H_4(g)} = 2292 - 2267 = 25 \text{ kJ mol}^{-1}$
Final Answer: The final answer is $25$
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