Calculate the enthalpy change for the process CCl4g- Cg-4Clgand calculate bond enthalpy of C-Cl in CCl4g - vapH- CCl4-30-5 kJ mol -1- - fH- CCl4-135-5 kJ mol -1- - aH- C-715-0 kJ mol -1- where - aH- is enthalpy of atomisation - aH- Cl2-242 kJ mol -1-

The given chemical reactions are: CCl_4(l)→ CCl_4(g) Δ_vap H^⊖=30.5 #160;kJ mol ^ 1 . C(g)+2Cl_2(g)→ CCl_4(g) Δ_f H^⊖= 135.5 #160;kJ mol ...

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Standard XII

Chemistry

Gibb's Energy and Nernst Equation

Calculate the...

Question

Calculate the enthalpy change for the process

$C C l_{4} (g) \to C (g) + 4 C l (g)$

and calculate bond enthalpy of C-Cl in $C C l_{4} (g)$

$Δ_{v a p} H^{⊖} (C C l_{4}) = 30.5$ kJ mol $^{- 1}$ .

$Δ_{f} H^{⊖} (C C l_{4}) = - 135.5$ kJ mol $^{- 1}$ .

$Δ_{a} H^{⊖} (C) = 715.0$ kJ mol $^{- 1}$ .

where $Δ_{a} H^{⊖}$ is enthalpy of atomisation

$Δ_{a} H^{⊖} (C l_{2}) = 242$ kJ mol $^{- 1}$ .

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Solution

The given chemical reactions are:

$C C l_{4} (l) \to C C l_{4} (g) Δ_{v a p} H^{⊖} = 30.5$ kJ mol $^{- 1}$ .

$C (g) + 2 C l_{2} (g) \to C C l_{4} (g) Δ_{f} H^{⊖} = - 135.5$ kJ mol $^{- 1}$ .
$C (s) \to C (g) Δ_{a} H^{⊖} = 715.0$ kJ mol $^{- 1}$ .
$C l_{2} (g) \to 2 C l (g) Δ_{a} H^{⊖} = 242$ kJ mol $^{- 1}$ .

Now for calculating enthalpy change for the process $C C l_{4} (g) \to C (g) + 4 C l (g)$ , we have to follow following algebric process

Equation $(2) + 2 \times$ Equation $(3) -$ Equation $(1) -$ Equation $(4)$

$Δ H = Δ_{a} H^{⊖} + 2 Δ_{a} H^{⊖} C l_{2} - Δ_{v a p} H^{⊖} - Δ_{f} H^{⊖}$

$= 715 + 2 \times 242 - 30.5 - (- 135.5)$

$Δ H = 1304$ kJ mol $^{- 1}$ .

Now,

The bond enthalpy of $C - C l$ in $C C l_{4} = \frac{1304}{4} = 326$ kJ mol $^{- 1}$ .

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Similar questions

Calculate the enthalpy change for the process

CCl₄₍_g₎ → C₍_g₎ + 4Cl₍_g₎

and calculate bond enthalpy of C–Cl in CCl₄₍_g_).

Δ_vapH^θ (CCl₄) = 30.5 kJ mol^–1.

Δ_fH^θ (CCl₄) = –135.5 kJ mol^–1.

Δ_aH^θ (C) = 715.0 kJ mol^–1, where Δ_aH^θ is enthalpy of atomisation

Δ_aH^θ (Cl₂) = 242 kJ mol^–1

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Enthalpy of sublimation of graphite =

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Q. Calculate the enthalpy of vaporisation (in

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Standard enthalpy of combustion of $C H_{4}$ is $- 890 k J m o l^{- 1}$ and standard enthalpy of vaporisation of water is 40.5 kJ $m o l^{- 1}$ . Calculate the enthalpy change for the reaction :
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Calculate the enthalpy change for the process CCl4(g)→C(g)+4Cl(g)and calculate bond enthalpy of C-Cl in CCl4(g)ΔvapH⊖(CCl4)=30.5 kJ mol−1. ΔfH⊖(CCl4)=−135.5 kJ mol−1. ΔaH⊖(C)=715.0 kJ mol−1. where ΔaH⊖ is enthalpy of atomisation ΔaH⊖(Cl2)=242 kJ mol−1.