If enthalpy of hydrogenation of ( C )6( H )6( l ) into ( C )6( H )12( l ) is -205 k J and resonance energy of ( C )6( H )6( l ) is -152 kJ mol -1 then enthalpy of hydrogenation of (l) to ( C )6( H )12( l ) is? [Assume enthalpies of vapourisation of all liquids involved are equal].

Question

If enthalpy of hydrogenation of ( C )6( H )6( l ) into ( C )6( H )12( l ) is -205 k J and resonance energy of ( C )6( H )6( l ) is -152 kJ mol -1 then enthalpy of hydrogenation of <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/c-zob7jkbvdcta4dmq.jpg" /> (l) to ( C )6( H )12( l ) is? [Assume enthalpies of vapourisation of all liquids involved are equal]. (A) - 535.5 kJ/mol (B) - 238 kJ/mol (C) - 357 kJ/mol (D) - 119 kJ/mol

Tardigrade Admin · Accepted Answer

C 6H 6(l)+3H 2( g ) arrow C 6 H 12(l) Δ H hyd ( C6 H 6) l=-205 kJ Benzene on reactant side so add resonance energy of benzene with sign. Heat evolved for hydrogenation of three double bonds =-205-152=-357 kJ mol -1 Δ H hyd °=(-357/3)=-119 kJ / mole

Q. If enthalpy of hydrogenation of (C)6​(H)6​(l) into (C)6​(H)12​(l) is −205kJ and resonance energy of (C)6​(H)6​(l) is −152kJmol−1 then enthalpy of hydrogenation of (l) to (C)6​(H)12​(l) is? [Assume enthalpies of vapourisation of all liquids involved are equal].

- 535.5 kJ/mol

- 238 kJ/mol

- 357 kJ/mol

- 119 kJ/mol

C6​H6​(l)+3H2​(g)→C6​H12​(l) ΔHhyd(C6​H6​)l​=−205kJ Benzene on reactant side so add resonance energy of benzene with sign. Heat evolved for hydrogenation of three double bonds =−205−152=−357kJmol−1 ΔHhyd∘​=3−357​=−119kJ/ mole