he enthalpy change for the reaction, CCl4(g) → C(g)+4Cl(g) is Given: Δ textvapH°(CCl4)=30.5 KJ mol-1, ΔfH°(CCl4)=-135.5 KJ mol-1, ΔaH°(C)=715.0 kJ mol-1, ΔaH° (Cl2)=242 KJ mol-1 where ΔaH° is enthalpy of atomisation

Question

he enthalpy change for the reaction, CCl4(g) → C(g)+4Cl(g) is Given: Δ textvapH°(CCl4)=30.5 KJ mol-1, ΔfH°(CCl4)=-135.5 KJ mol-1, ΔaH°(C)=715.0 kJ mol-1, ΔaH° (Cl2)=242 KJ mol-1 where ΔaH° is enthalpy of atomisation (A) 1200 kJ mol-1 (B) 500 kJ mol-1 (C) 1304 kJ mol-1 (D) 1500 kJ mol-1

Tardigrade Admin · Accepted Answer

Given: (i) CCl4(l) arrow CCl4(g), Δ textvap H=30.5 kJ mol-1 (ii) C(s)+2Cl2(g) arrow CCl4(l), Δf H°=-135.5 KJ mol-1 (iii) C(s)arrow C(g), ΔaH°=715.0 KJ mol-1 (iv) Cl2(g)arrow2Cl(g), ΔaH°=242 KJ mol-1 Required equation is: CCl4(g) → C(g)+4Cl(g); Δ H= ? From Hess’s law, eqn.(iii) + 2 × eqn.(iv) - eqn.(i) - eqn.(ii) gives required equation: ∴ Δ H = 715.0+2(242)-30.5-(-135.5) =1304 kJ mol-1

$1200 k J m o l^{- 1}$

$500 k J m o l^{- 1}$

$1304 k J m o l^{- 1}$

$1500 k J m o l^{- 1}$

Q. he enthalpy change for the reaction, CCl4(g)​→C(g)​+4Cl(g)​ is Given: Δvap​H∘(CCl4​)=30.5KJmol−1,Δf​H∘(CCl4​)=−135.5KJmol−1, Δa​H∘(C)=715.0kJmol−1,Δa​H∘(Cl2​)=242KJmol−1 where Δa​H∘ is enthalpy of atomisation

1200kJmol−1

500kJmol−1

1304kJmol−1

1500kJmol−1

$1200 k J m o l^{- 1}$

$500 k J m o l^{- 1}$

$1304 k J m o l^{- 1}$

$1500 k J m o l^{- 1}$