If enthalpy of hydrogenation of mathrmC6 mathrmH6( mathrml) into mathrmC6 mathrmH12 (l) is -205 mathrm~kJ and resonance energy of mathrmC6 mathrmH6( mathrml) is -152 mathrm~kJ mathrm~mol-1 then enthalpy of (l) to ( C )6( H )12( l ) is? [Assume enthalpiies of vapourisation of all liquids involved are equal].

Question

If enthalpy of hydrogenation of mathrmC6 mathrmH6( mathrml) into mathrmC6 mathrmH12 (l) is -205 mathrm~kJ and resonance energy of mathrmC6 mathrmH6( mathrml) is -152 mathrm~kJ mathrm~mol-1 then enthalpy of <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/c-6fq8xdze7aks.png" /> (l) to ( C )6( H )12( l ) is? [Assume enthalpiies 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

beginarrayc mathrmC6 mathrmH6(l)+3 mathrmH2( mathrm~g) longrightarrow mathrmC6 mathrmH12(l) Δ mathrmH mathrmhyd( mathrmC6 mathrmH6) l°=-205 mathrm~kJ endarray 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 texthyd °=(-357/3)=-119 kJ / mole

Q. If enthalpy of hydrogenation of C6​H6​(l) into C6​H12​ (l) is −205 kJ and resonance energy of C6​H6​(l) is −152 kJ mol−1 then enthalpy of (l) to (C)6​(H)12​(l) is? [Assume enthalpiies 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∘​=−205 kJ​ 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

Q. If enthalpy of hydrogenation of $C_{6} H_{6} (l)$ into $C_{6} H_{12}$ (l) is $- 205 kJ$ and resonance energy of $C_{6} H_{6} (l)$ is $- 152 kJ mol^{- 1}$ then enthalpy of

(l) to $(C)_{6} (H)_{12} (l)$ is?
[Assume enthalpiies of vapourisation of all liquids involved are equal].