Determine C–C and C–H bond enthalpy (in kJ/mol) Given: _ f H ^ ( C _ 2 H _ 6 , ~g ) = –85…
Chemistry · JEE Advanced · NTA Exams — Chemical Thermodynamics
Determine C–C and C–H bond enthalpy (in kJ/mol)
Given: \(\Delta_{f} \mathrm{H}^{\circ}\left(\mathrm{C}_{2} \mathrm{H}_{6}, \mathrm{~g}\right)\)= –85 kJ/mol
\(\Delta_{f} \mathrm{H}^{\circ}\left(\mathrm{C}_{3} \mathrm{H}_{8}, \mathrm{~g}\right)\)= –104 kJ/mol
\(\Delta_{\mathrm{sub}} \mathrm{H}^{\circ}(\mathrm{C}, \mathrm{~s})\)=718 kJ/mol \(\text { B.E. }(\mathrm{H}-\mathrm{H})\)= 436 kJ/mol
Given: \(\Delta_{f} \mathrm{H}^{\circ}\left(\mathrm{C}_{2} \mathrm{H}_{6}, \mathrm{~g}\right)\)= –85 kJ/mol
\(\Delta_{f} \mathrm{H}^{\circ}\left(\mathrm{C}_{3} \mathrm{H}_{8}, \mathrm{~g}\right)\)= –104 kJ/mol
\(\Delta_{\mathrm{sub}} \mathrm{H}^{\circ}(\mathrm{C}, \mathrm{~s})\)=718 kJ/mol \(\text { B.E. }(\mathrm{H}-\mathrm{H})\)= 436 kJ/mol
- 414, 345
- 345, 414
- 287, 404.5
- None of these
Answer
(B) 345, 414
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