For the reaction CCl₄(g) → C(g) + 4Cl(g), if Δ_a H°(C) = 715.0 kJ mol⁻¹ and Δ_a H°(Cl₂) = 242 kJ mol

Q: What is the correct answer for CBSE11?

The correct answer is A. The reaction CCl₄(g) → C(g) + 4Cl(g) represents the atomization of CCl₄. The enthalpy change for this process is the sum of the energies required to break all the C-Cl bonds. There are 4 C-Cl bonds in CCl₄. So, 4 * (C-Cl bond enthalpy) = Δ_a H°(CCl₄). The problem provides Δ_a H°(C) and Δ_a H°(Cl₂), but not Δ_a H°(CCl₄) directly. However, the question asks for the bond enthalpy of a C-Cl bond, impl

Q: For the reaction CCl₄(g) → C(g) + 4Cl(g), if Δ_a H°(C) = 715.0 kJ mol⁻¹ and Δ_a H°(Cl₂) = 242 kJ mol⁻¹, what is the bond enthalpy of a C-Cl bond in CCl₄(g)? (Assume Δ_vap H°(CCl₄) = 30.5 kJ mol⁻¹ and…

The correct answer is A. The reaction CCl₄(g) → C(g) + 4Cl(g) represents the atomization of CCl₄. The enthalpy change for this process is the sum of the energies required to break all the C-Cl bonds. There are 4 C-Cl bonds in CCl₄. So, 4 * (C-Cl bond enthalpy) = Δ_a H°(CCl₄). The problem provides Δ_a H°(C) and Δ_a H°(Cl₂), but not Δ_a H°(CCl₄) directly. However, the question asks for the bond enthalpy of a C-Cl bond, impl

Q: What are the options for this CBSE11 question?

A. 326.5 kJ mol⁻¹ · B. 1306 kJ mol⁻¹ · C. 163.25 kJ mol⁻¹ · D. 653 kJ mol⁻¹

Question

For the reaction CCl₄(g) → C(g) + 4Cl(g), if Δ_a H°(C) = 715.0 kJ mol⁻¹ and Δ_a H°(Cl₂) = 242 kJ mol⁻¹, what is the bond enthalpy of a C-Cl bond in CCl₄(g)? (Assume Δ_vap H°(CCl₄) = 30.5 kJ mol⁻¹ and Δ_f H°(CCl₄) = -135.5 kJ mol⁻¹ are also given, but not directly needed for bond enthalpy calculation from atomization).

FirstInTest · Accepted Answer

The reaction CCl₄(g) → C(g) + 4Cl(g) represents the atomization of CCl₄. The enthalpy change for this process is the sum of the energies required to break all the C-Cl bonds. There are 4 C-Cl bonds in CCl₄. So, 4 * (C-Cl bond enthalpy) = Δ_a H°(CCl₄). The problem provides Δ_a H°(C) and Δ_a H°(Cl₂), but not Δ_a H°(CCl₄) directly. However, the question asks for the bond enthalpy of a C-Cl bond, implying the atomization enthalpy of CCl₄ is needed. If we assume the given reaction CCl₄(g) → C(g) + 4Cl(g) is the atomization process, then the total energy to break all bonds is the sum of atomization energies of C and Cl. This is a bit ambiguous as the question is phrased. Let's re-read. 'Calculate the enthalpy change for the process CCl₄(g) → C(g) + 4 Cl(g) and calculate bond enthalpy of C – Cl in CCl₄(g)'. This means we need to calculate the enthalpy change for the given reaction first. The enthalpy change for this reaction (atomization of CCl₄) would be Δ_r H° = Δ_f H°(C,g) + 4 * Δ_f H°(Cl,g) - Δ_f H°(CCl₄,g). We are given Δ_a H°(C) = 715.0 kJ mol⁻¹ (which is Δ_f H°(C,g)) and Δ_a H°(Cl₂) = 242 kJ mol⁻¹ (which is for Cl₂ → 2Cl). So Δ_f H°(Cl,g) = 242/2 = 121 kJ mol⁻¹. We need Δ_f H°(CCl₄,g). We are given Δ_f H°(CCl₄,l) = -135.5 kJ mol⁻¹ and Δ_vap H°(CCl₄) = 30.5 kJ mol⁻¹. So, Δ_f H°(CCl₄,g) = Δ_f H°(CCl₄,l) + Δ_vap H°(CCl₄) = -135.5 + 30.5 = -105 kJ mol⁻¹. Now, Δ_r H° (for CCl₄(g) → C(g) + 4Cl(g)) = 715.0 + 4*(121) - (-105) = 715.0 + 484 + 105 = 1304 kJ mol⁻¹. This is the total energy to break 4 C-Cl bonds. So, bond enthalpy of C-Cl = 1304 / 4 = 326 kJ mol⁻¹.

Answer

326.5 kJ mol⁻¹

Answer

1306 kJ mol⁻¹

Answer

163.25 kJ mol⁻¹

Answer

653 kJ mol⁻¹

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