Given the following data: Δ f H 0 ( C H 4 , g ) = –80 k J m o l − 1 Δ f H 0 ( C 2 H 6 , g ) = –85 k J m o l − 1 Δ f H 0 ( C 2 H 4 , g ) = 50 k J m o l − 1 Enthalpy of sublimation of ca

Question

Given the following data: Δ f ​ H 0 ( C H 4 ​ , g ) = –80 k J m o l − 1 Δ f ​ H 0 ( C 2 ​ H 6 ​ , g ) = –85 k J m o l − 1 Δ f ​ H 0 ( C 2 ​ H 4 ​ , g ) = 50 k J m o l − 1 Enthalpy of sublimation of carbon (graphite) = 720 k J m o l − 1 Enthalpy of atomization of H = 215 k J m o l − 1 List - I List - II (I) Bond enthalpy of C − H bond (per mole) (P) 325 kJ (II) Bond enthalpy of C − C bond (per mole) (Q) 600 kJ (III) Bond enthalpy of C = C bond (per mole) (R) 415 kJ (IV) Δ H 0 for the reaction : n C 2 ​ H 4 ​ ( g ) → ( C H 2 ​ − C H 2 ​ − ) n ​ ( g ) (for 10 moles of ethene) (T) 650 kJ (U) 375 kJ Which of the following options has the correct combination considering list-I and list-II

TestHub · Accepted Answer

Given: Δ f ​ H 0 ( C H 4 ​ ) ​ = − 80 Δ f ​ H 0 ( C 2 ​ H 6 ​ ) = − 85 Enthalpy of sublimation of carbon (graphite) = 720 k J m o l − 1 Enthalpy of atomization of H = 215 k J m o l − 1 Using: Δ H = ∑ B o n d e n th a lp h y ∘ f b o n d s b ro k e n − ∑ B o n d e n th a lp h y ∘ f b o n d s f or m e d For methane: C H 4 ​ ( g ) → C ( g ) + 4 H ​ ( g ) ) Hence, \displaylines 4 D ( C − H ) = 1660 D ( C − H ) = 4 1660 ​ = 415 k J m o l − 1 For ethane: C 2 ​ H 6 ​ → 2 C + 6 H ​ \displaylines D ( C − C ) + 2490 = 2815 D ( C − C ) = 2815 − 2490 = 325 k J m o l − 1

List - I	List - II
(I) Bond enthalpy of $C - H$ bond (per mole)	(P) 325 kJ
(II) Bond enthalpy of $C - C$ bond (per mole)	(Q) 600 kJ
(III) Bond enthalpy of $C = C$ bond (per mole)	(R) 415 kJ
(IV) $Δ H^{0}$ for the reaction : $n C_{2} H_{4} (g) \to (C H_{2} - C H_{2} -)_{n} (g)$ (for 10 moles of ethene)	(T) 650 kJ
	(U) 375 kJ

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